Naked Science Forum
On the Lighter Side => New Theories => Topic started by: butchmurray on 06/06/2013 05:05:00
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LENGTH CONTRACTION AND TIME DILATION CONTRADICT THE CONSTANCY OF THE SPEED OF LIGHT - MMXIII
By
Thorntone E. ‘Butch’ Murray
Houston, Texas, USA
June 4, 2013
CONTENTION:
This contention is in strict adherence to and based exclusively on principles presented in the 1905 Theory of Special Relativity (SR) by Albert Einstein. The following mathematical analysis of an ordinary, unremarkable circumstance proves that time dilation and length contraction as presented in SR contradict a fundamental assumption of SR that the speed of light is constant and the same for all observers.
FOR THESE PURPOSES:
There are two identical circular clocks each with a pointer and there are three identical light clocks each with a light pulse. A light clock is a particular type of light path. The length of a measuring rod, the length of a light clock and the length of a light path are analogous. When light clocks are judged to have the same length, the light pulses in those light clocks are judged to have the same round trip distance. When the length of a light clock is judged contracted by a factor, the round trip distance for the light pulse in that light clock is judged contracted by that same factor. The amount of time for the pointer in a circular clock to complete one revolution is the amount of time for light pulses in light clocks in the same frame to complete one round trip. Frame K and frame K’ are inertial frames. Frame K’ is in motion relative to frame K.
IN ACCORDANCE WITH LENGTH CONTRACTION
AS PRESENTED IN SR:
The proper length of a measuring rod and so the proper length of a light clock is its length judged from within the frame it occupies. The three light clocks are the same proper length. The round trip distance for a light pulse in a light clock of proper length is d. Judged from frame K the length of a measuring rod and so the length of a light clock oriented perpendicular to the direction of motion in frame K’ is its proper length. Then, judged from frame K the round trip distance for a light pulse in a light clock in frame K and the round trip distance for a light pulse in a light clock oriented perpendicular to the direction of motion in frame K’ are both d. Judged from frame K the length of a measuring rod and so the length of a light clock oriented in the direction of motion in frame K’ is its proper length contracted by the length contraction factor sqrt(1-(v*v/c*c)). The round trip distance for a light pulse in a light clock judged as its proper length contracted by the length contraction factor sqrt(1-(v*v/c*c)) is d*sqrt(1-(v*v/c*c)). Then, judged from frame K the round trip distance for a light pulse in a light clock oriented in the direction of motion in frame K’ is d*sqrt(1-(v*v/c*c)).
IN ACCORDANCE WITH TIME DILATION
AS PRESENTED IN SR:
Judged from frame K when K’ is in motion relative to frame K time in K’ advances at a slower rate than time in frame K. Judged from frame K time in frame K advances at a faster rate than time in K’. Judged from frame K time in frame K advances at a faster rate than time in K’ by the factor 1/sqrt(1-(v*v/c*c)). Then judged from frame K the pointer in the circular clock in K advances at a faster rate than the pointer in the identical circular clock in frame K’ by the factor 1/sqrt(1-(v*v/c*c)). Judged from K the amount of time for the pointer in the circular clock in K’ to complete one revolution is the amount of time for the pointer in the identical circular clock in frame K to complete 1/sqrt(1-(v*v/c*c)) revolutions. Judged from frame K t is the amount of time for the pointer in the circular clock in frame K to complete one revolution. Then judged from K the amount of time for the pointer in the circular clock in frame K’ to complete one revolution is t/sqrt(1-(v*v/c*c)).
THE LIGHT CLOCK IN FRAME K:
One of the previously described circular clocks and one of the light clocks are in frame K. Judged from K that light clock is its proper length. Then, judged from frame K the round trip distance for the light pulse in that light clock in frame K is d. Judged from K the pointer in the circular clock in frame K completes one revolution in amount of time t. The amount of time for the pointer in a circular clock to complete one revolution is the amount of time for a light pulse to complete one round trip in light clocks in the same frame. Then, judged from K the light pulse in the light clock in frame K completes one round trip in that light clock in amount of time t.
For these purposes, v is the speed of the light pulse, the speed of light c. The amount of time for the light pulse to complete the round trip is t. The round trip distance for the light pulse is d. All is judged from frame K.
The speed formula
v=d/t
Substitute c for v
c=d/t
For these purposes, judged from frame K the expression (d/t) represents the speed of the light pulse thus the speed of light in the light clock in frame K.
BOTH LIGHT CLOCKS IN K’:
The remaining circular clock and two light clocks are in frame K’. The amount of time for a light pulse to complete one round trip in the light clocks in frame K’ is the amount of time for the pointer in the circular clock in frame K’ to complete one revolution. Judged from K the amount of time for the pointer in the circular clock in frame K’ to complete one revolution is the amount of time for the pointer in the circular clock in K to complete 1/sqrt(1-(v*v/c*c)) revolutions. Judged from K the amount of time for a light pulse to complete one round trip in the light clocks in frame K’ is the amount of time for the pointer in the circular clock in K to complete 1/sqrt(1-(v*v/c*c)) revolutions. Judged from K the amount of time for the pointer in the circular clock in K to complete 1/sqrt(1-(v*v/c*c)) revolutions is t/sqrt(1-(v*v/c*c)). Therefore, judged from K the amount of time for a light pulse to complete one round trip in the light clocks in frame K’ is t/sqrt(1-(v*v/c*c)).
THE LIGHT CLOCK PERPENDICULAR TO
THE DIRECTION OF MOTION IN K’:
One of the light clocks in frame K’ is oriented perpendicular to the direction of motion. Judged from frame K the length of the light clock that is perpendicular to the direction of motion in frame K’ is its proper length. Then, judged from frame K the round trip distance for the light pulse in that light clock is d. Judged from frame K the amount of time for the light pulse in each light clock in K’ complete one round trip is t/sqrt(1-(v*v/c*c)). Judged from K the round trip distance for the light pulse in the light clock in frame K’ that is oriented perpendicular to the direction of motion is d.
For these purposes, v is the speed of the light pulse, the speed of light c. The amount of time for the light pulse to complete the round trip is t/sqrt(1-(v*v/c*c)). The round trip distance for the light pulse is d. All is judged from frame K.
The speed formula
v=d/t
Substitute c for v
c=d/t
Substitute t/sqrt(1-(v*v/c*c)) for t
c=d/t/sqrt(1-(v*v/c*c))
Simplify
c=(d/t)*sqrt(1-(v*v/c*c))
For these purposes, judged from frame K the expression (d/t)*sqrt(1-(v*v/c*c)) represents the speed of the light pulse thus the speed of light in the light clock that is perpendicular to the direction of motion in frame K’.
BOTH LIGHT CLOCKS IN K’: (Intentionally repeated)
Previously established under this heading, judged from frame K the amount of time for a light pulse to complete one round trip in the light clocks in frame K’ is t/sqrt(1-(v*v/c*c)).
THE LIGHT CLOCK IN
THE DIRECTION OF MOTION IN K’:
The remaining light clock in frame K’ is oriented in the direction of motion. Judged from frame K the length of the light clock that is in the direction of motion in frame K’ is its proper length contracted by the length contraction factor sqrt(1-(v*v/c*c)). Then, judged from frame K the round trip distance for the light pulse in that light clock is d*sqrt(1-(v*v/c*c)). Judged from frame K the amount of time for the light pulse in each light clock in K’ to complete one round trip is t/sqrt(1-(v*v/c*c)). Judged from K the round trip distance for the light pulse in the light clock in frame K’ that is oriented in the direction of motion is d*sqrt(1-(v*v/c*c)).
For these purposes, v is the speed of the light pulse, the speed of light c. The amount of time for the light pulse to complete the round trip is t/sqrt(1-(v*v/c*c)). The round trip distance for the light pulse is d*sqrt(1-(v*v/c*c)). All is judged from frame K.
The speed formula
v=d/t
Substitute c for v
c= d/t
Substitute t/sqrt(1-(v*v/c*c)) for t
c=d/t/sqrt(1-(v*v/c*c))
Simplify
c=(d/t)*sqrt(1-(v*v/c*c))
Substitute d*sqrt(1-(v*v/c*c)) for d
c=(d*sqrt(1-(v*v/c*c))/t)*sqrt(1-(v*v/c*c))
Simplify
c=(d/t)*sqrt(1-(v*v/c*c))*sqrt(1-(v*v/c*c))
Simplify further
c=(d/t)*(1-(v*v/c*c))
For these purposes, judged from frame K the expression (d/t)*(1-(v*v/c*c)) represents the speed of the light pulse thus the speed of light in the light clock that is in the direction of motion in frame K’.
SUMMARY:
Judged by an observer in frame K with frame K’ in motion relative to frame K at a relative speed greater than zero and less than c the speed of light in each of the three light clocks is unique.
Judged by the observer in frame K the speed of light in the light clock in inertial frame K:
c= d/t
Judged by the observer in frame K the speed of light in the light clock perpendicular to the direction of motion in inertial frame K’:
c=(d/t)*sqrt(1-(v*v/c*c))
Judged by the observer in frame K the speed of light in the light clock in the direction of motion in inertial frame K’:
c=(d/t)*(1-(v*v/c*c))
The speed of light in the three light clocks is not constant and the same judged by an observer in frame K.
CONCLUSION:
This mathematical analysis of an ordinary, unremarkable circumstance proves that time dilation and length contraction as presented in SR contradict a fundamental assumption of SR that the speed of light is constant and the same for all observers.
Butchmurray
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I don't think anyone's going to be able to find the time to try to understand all that. Perhaps some diagrams would help make it more intelligible. There's a tool for drawing them here which can be accessed by clicking on "Create a new diagram", just under the box where you type your message in (assuming you aren't using the "Quick reply" one).
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Hi David.
Your response to my message was quicker than I anticipated.
You make a good point. I’ll post a synopsis in the next few days.
Thanks for the insight.
Butch
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I apologize for my prolonged absence.
Butch
This explanation by example should make the concept easier to understand. The math is basic.
The contention uses only the principles as set forth in The Theory of Special Relativity (SR) and the speed formula, v=d/t. It proves mathematically that length contraction and time dilation as set forth by SR contradict the fundamental assumption of SR that the speed of light is constant and the same for all observers. Judged from within the frame of the observer the speed of light is 300,000 km/s. The same observer judges the speed of light perpendicular to the direction of motion to be 150,000 km/s and the speed of light in the direction of motion to be 75,000 km/s in an inertial frame in relative motion at the at the speed .866c.
FOR THESE PURPOSES:
A light clock is an imaginary construction that consists of two reflectors with a light pulse continuously reflected from one reflector to the other.
Judged from within the inertial frame it occupies the length of a light clock is its proper length.
For a light clock judged as its proper length, the round trip distance for the light pulse in that light clock is 300,000 km.
For a light clock judged as its proper length contracted by the length contraction factor, the round trip distance for the light pulse in that light clock is judged as 300,000 km contracted by that same factor.
There are 2 identical circular clocks and 3 identical light clocks.
The light pulse in each light clock completes 1 round trip in the time it takes for the pointer in the circular clock in that same frame to complete 1 revolution.
Judged from within the inertial frame it occupies the pointer in a circular clock completes 1 revolution in 1 second.
1 circular clock and 1 light clock are in frame K with the observer.
1 circular clock and 2 light clocks are in frame K’.
Inertial frame K’ is in motion relative to inertial frame K at the speed .866c.
THE LIGHT CLOCK IN FRAME K:
Judged by the observer in frame K the length of the light clock is its proper length. Then judged by the observer in frame K the round trip distance for the light pulse in that light clock is 300,000 km. Judged by the observer in frame K the amount of time for that light pulse to complete 1 round trip and the time for the pointer in the circular clock to complete 1 revolution is identical, 1 second.
The speed formula
v=d/t
Judged by the observer in frame K
d= 300,000 km - the round trip distance for the light pulse
t= s - 1 second
v= c the speed of the light pulse
Substitute 300,000 km for d, s for t, c for v
c=300,000 km/s
Judged by the observer in frame K the speed of light in the light clock in frame K is 300,000 km/s.
Inertial frame K is at relative rest, inertial frame K’ is in relative motion at .866c.
Per SR relative to time in frame K time in frame K’ is dilated/slower.
Per SR for the relative speed .866c the time dilation factor is 2.
Per SR judged by the observer in frame K length perpendicular to the direction of motion in frame K’ is the proper length.
Per SR judged by the observer in frame K length in the direction of motion in frame K’ is the proper length factored by the length contraction factor.
Per SR for the relative speed .866c the length contraction factor is .5.
FOR BOTH LIGHT CLOCKS IN K’:
The light pulse in each of the light clocks in K’ complete 1 round trip and the pointer in the circular clock in K’ completes 1 revolution in the same amount of time. As the time dilation factor is 2, judged by the observer in frame K the pointer in the circular clock in K’ completes 1 revolution and the pointer in the circular clock in frame K completes 2 revolutions in the same amount of time. Then judged by the observer in frame K the light pulse in each of the light clocks in K’ complete 1 round trip and the pointer in the circular clock in frame K completes 2 revolutions in the same amount of time. Judged by the observer in frame K the pointer in the circular clock in frame K completes 1 revolution in 1 second. Then judged by the observer in frame K the pointer in the circular clock in frame K completes 2 revolutions in 2 seconds. Then judged by the observer in frame K the light pulse in each of the light clocks in K’ complete 1 round trip in 2 seconds.
THE LIGHT CLOCK PERPENDICULAR TO THE DIRECTION OF MOTION IN FRAME K’:
Judged by the observer in frame K the length of the light clock that is perpendicular to the direction of motion in frame K’ is the proper length. Then judged by the observer in frame K the round trip distance for the light pulse in that light clock is 300,000 km. Stated in “FOR BOTH LIGHT CLOCKS IN K’ ”, judged by the observer in frame K the light pulse in each of the light clocks in K’ complete 1 round trip in 2 seconds.
The speed formula
v=d/t
Judged by the observer in frame K
d= 300,000 km - the round trip distance for the light pulse
t= 2s - two seconds
v= c the speed of the light pulse
Substitute 300,000 km for d, 2s for t, c for v
c=300,000 km/2s
Simplify
c=150,000 km/s
Judged by the observer in frame K the speed of light in the light clock perpendicular to the direction of motion in frame K’ is 150,000 km/s.
THE LIGHT CLOCK IN THE DIRECTION OF MOTION IN FRAME K’:
Judged by the observer in frame K the length of the light clock that is in the direction of motion in frame K’ is the proper length factored by the length contraction factor .5. Then judged by the observer in frame K the round trip distance for the light pulse in that light clock is 300,000 km factored by .5. Stated in “FOR BOTH LIGHT CLOCKS IN K’ ”, judged by the observer in frame K the light pulse in each of the light clocks in K’ complete 1 round trip in 2 seconds.
The speed formula
v=d/t
Judged by the observer in frame K
d= .5*300,000 km the round trip distance for the light pulse factored by .5
t= 2s - two seconds
v= c the speed of the light pulse
Substitute .5*300,000 km for d, 2s for t, c for v
c=.5*300,000 km/2s
Simplify
c=150,000 km/2s
Simplify further
c=75,000 km/s
Judged by the observer in frame K the speed of light in the light clock in the direction of motion in frame K’ is 75,000 km/s.
Butch
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I'm sorry, but I've cracked. I just can't load all that stuff into my head any more. Maybe someone else can have a go.
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David,
Please don’t crack now!
Is this better?
Judged by one observer a light pulse, therefore, light has 3 distinct speeds.
Everything here is judged by one observer in the inertial frame that is at relative rest.
WITHIN THE FRAME OF THE OBSERVER:
Orientation of light path – does not apply
Length – proper length
Time – proper time
Length of the light path – the proper length, 299,792,548 meters (300,000 km)
Time for the light pulse to traverse the light path – 1 second proper time
Speed of the light pulse in the light path – 300,000 km/1 second
300,000 km/s – the proper speed of light judged by the observer
Speed of relatively moving frame - .866c
Time dilation factor for .866c – 2
Length contraction factor for .866c - .5
THE RELATIVELY MOVING FRAME:
LIGHT PATH PERPENDICULAR TO THE DIRECTION OF MOTION:
Length – proper length
Length of the light path – the proper length 300,000 km
Time for moving frame – proper time is factored by 2, the time dilation factor
Time dilation factor 2 * 1 second proper time = 2 seconds dilated time
Time for the light pulse to traverse the light path – 2 seconds dilated time
Speed of the light pulse in the light path – 300,000 km/2 seconds
150,000 km/s – Speed of light judged by the observer (at relative rest)
LIGHT PATH IN THE DIRECTION OF MOTION:
Length – proper length is factored by .5, the length contraction factor
Length contraction factor .5 * the proper length 300,000 km = 150,000 km
Length of the light path – the contracted length 150,000 km
Time for moving frame – proper time is factored by 2, the time dilation factor
Time dilation factor 2 * 1 second proper time = 2 seconds dilated time
Time for the light pulse to traverse the light path – 2 seconds dilated time
Speed of the light pulse in the light path – 150,000 km/2 seconds
75,000 km/s – Speed of light judged by the observer (at relative rest)
JUDGED BY THE OBSERVER AT RELATIVE REST:
300,000 km/s is the proper speed of light in the frame of the observer
150,000 km/s is the speed of light in a light path perpendicular to the direction of motion in a frame in relative motion at .866c
75,000 km/s is the speed of light in a light path in the direction of motion in a frame in relative motion at .866c
Butch
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Sorry, but I just can't do it any more. It's still all presented in a form that I simply can't get my mind to work with. I'm sure your argument could be presented in a form where the central part of it comes up front and where it's described clearly to avoid unnecessary confusion, but as it stands it's asking the reader to do ten times as much work as should be necessary. Experience also tells me that what you have above is likely to be nothing more than a more convoluted version of your earlier attempts to prove something where it didn't appear to work and you didn't appear to be able to see that it didn't work (or maybe where it did work and I just couldn't see it, in which case I'm not up to the task anyway), so I simply can't motivate myself to work with the new version unless it's presented enormously better than any of the past versions (which were always far too hard to follow comfortably). It's just got to the point where I can't get my mind to work on it any more, and it's beginning to turn into a mental block now which means that even if you do manage to present it properly I'm going to have a hard time getting up the energy to attempt to work with it, but it would at least make it more likely that someone else would give it a go. [I still think diagrams would help.]
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SYNOPSIS:
Speed = distance divided by time, d/t e.g. 300,000 kilometers/1 second or simply 300,000 km/s
IN ACCORDANCE WITH THE THEORY OF SPECIAL RELATIVITY (SR):
Judged by an observer at relative rest the speed of a light pulse (the speed of light) in a light path in the OBSERVER’S FRAME is defined as the proper length of the light path (its length in the observer’s frame) divided by the proper time (time in the observer’s frame):
- proper length/proper time
Judged by that same observer at relative rest, when a light path in a FRAME IN MOTION relative to that observer is PERPENDICULAR TO THE DIRECTION OF MOTION the speed of the light pulse is defined as the proper length of the light path divided by the product of the proper time multiplied by the time dilation factor:
- proper length/proper time * time dilation factor
Again, judged by that same observer at relative rest, when that light path in that same FRAME IN MOTION is IN THE DIRECTION OF MOTION, the speed of the light pulse is defined as the proper length of the light path multiplied by the length contraction factor divided by the product of the proper time multiplied by the time dilation factor:
- proper length * length contraction factor/proper time * time dilation factor
Then in accordance with SR, for that one observer light has 3 distinct speeds.
- proper length/proper time
- proper length/proper time * time dilation factor
- proper length * length contraction factor/proper time * time dilation factor
HOWEVER, according to SR the speed of light is constant and the same for all observers.
Butch
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Frames K and K' have their origins O and O' coincident at time t=t'=0. Let x be the direction of motion of K' with respect to K the speed being v. A ray of light leaves the origin and propagates in the vertical direction y with speed C, as seen by an observer at rest in the K' frame. We have C = d'/t', where d' is the distance traveled in the time t'. The distance d' could be represented by a vertical rod of length d' = O'A'.
Another ray of light leaves the origin and propagates in the vertical direction y with speed C, as seen by an observer at rest in the K frame. We have C = d/t, where d is the distance traveled in the time t. The distance d could be represented by a vertical rod of length d = OA.
The vertical distances traveled, d and d', are equal in both K and K'. Since C = d'/t' then t’=d’/C. Since C = d/t then t=d/C. Since d=d’ then t’=d/C and t=d/C therefore, t’=t.
Since t’=t, then time is equal in both frame K and frame K’.
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This is an unpolished version of the quantitative mathematical description of the contention. A significant modification was made.
Frames K and K' have their origins O and O' coincident at time t=t'=0. Let x be the direction of motion of K' with respect to K the speed being v. A ray of light leaves the origin and propagates in the vertical direction y with speed C, as seen by an observer at rest in the K frame. We have C = d/t, where d is the distance traveled in the time t. The distance d could be represented by a vertical rod of length d = OA. Then, as seen by an observer at rest in the K frame C = d/t=C
Another ray of light leaves the origin and propagates in the vertical direction y with speed C, as seen by an observer at rest in the K' frame. We have C = d'/t', where d' is the distance traveled in the time t'. The distance d' could be represented by a vertical rod of length d' = O'A'. When d’ is in the vertical direction y, d’=d. (Per SR, time in K’ is dilated relative to time in K.) So, t’=(t)(square root[1-(v²/C²)]). Then, as seen by an observer at rest in the K frame C = d'/t' = d/(t)(square root[1-(v²/C²)]) =C
An additional ray of light leaves the origin and propagates in the horizontal direction x with speed C, as seen by an observer at rest in the K' frame. We have C = d'/t', where d' is the distance traveled in the time t'. The distance d' could be represented by a horizontal rod of length d'. When d’ is in the horizontal direction x, d’=(d)(square root[1-(v²/C²)]). Time, t’=(t)(square root[1-(v²/C²)]). Then, as seen by an observer at rest in the K frame C = d'/t' = (d)(square root[1-(v²/C²)])/(t)(square root[1-(v²/C²)]) = d/t=C
As seen by an observer at rest in the K frame the speed of light is not constant:
C = d/t within frame K
C = d/(t)(square root[1-(v²/C²)]) in the vertical direction y in frame K’
C = d/t in the horizontal direction x in frame K’
Butch
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Dear Dr.
Is the proof of the inconsistency of the speed of light within SR clarified for you?
Here is something completely different:
The equation t’=(t)square root(1-v²/C²) is obtained from the equation (vt)² + (Ct’)² = (Ct)². The time dilation factor is 1/square root(1-v²/C²).
The equation L’=(L)square root(1-v²/C²) is sited as the equation from which the length contraction factor was obtained. The length contraction factor is: square root(1-v²/C²). The length contraction factor does not apply to length perpendicular to the direction of motion.
How is L’=(L)square root(1-v²/C²) mathematically obtained? How is its directionality mathematically justified? If either of these questions has no mathematically valid answer the length contraction factor, and therefore, length contraction are not scientifically valid. The most ardent supporters of SR would be forced to abandon it in that case.
Thank you,
Butch
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Pursuant to SR, for measuring rods of identical proper length that are perpendicular to the direction of motion, length is the same in all frames and the same for all observers. Pursuant to SR, the speed of light is equal in all frames and the same for all observers. Then, pursuant to the laws of physics, the time for light to propagate the length of measuring rods of identical proper length that are perpendicular to the direction of motion is the same in all frames and the same for all observers. Therefore, in accordance with the laws of physics, time is the same in all frames and the same for all observers.
Inertial frame K’ is in motion relative to frame K at a speed >0<C. Within frame K’ as seen by the observer in K’ light propagates length perpendicular to the direction of motion, d’, at speed C in the time t’. Within frame K as seen by the observer in K light propagates length perpendicular to the direction of motion, d, at speed C in the time t. Per SR, d’=d.
d=Ct
Solved for t --------------- t=d/C
d’=d
d’=Ct’
Substitute d for d’
d=Ct’
Solved for t’ -------------- t’=d/C
Then, ---------------------- t’=t
The time dilation factor and the length contraction factor are derived from the equation (vt)² + (Ct’)² = (Ct)² solved for t’.
(vt)² + (Ct’)² = (Ct)² --------------- Solve for t’
(Ct’)² = (Ct)² - (vt)²
C²t’² = t² (C² - v²)
t’² = t² (1- v²/ C²)
t’ = (t)square root(1- v²/ C²)------- Solved for t’
t’=t Proved previously
Substitute t for t’
t = (t)square root(1- v²/ C²)---Obviously, the formula is invalid.
The formula t’ = (t)square root(1- v²/ C²) and everything derived from it are invalid. The formula is dependent on t’<t. It is now proved t’=t. Therefore:
t=t’/square root(1-v²/C²)--- Time Dilation – is invalid
L’(x)=L(x)square root(1-v²/C²)---Length Contraction – is invalid
1/square root(1-v²/C²)--- gamma – is invalid
September 18, 2013
Thorntone Errick ‘Butch’ Murray
butchmurray
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To save my having to wade through the maths without a diagram, let's cut to the chase. What experimental result does your theory predict? How is this different from a conventional SR prediction? What was the actual result of the experiment?
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LENGTH CONTRACTION AND TIME DILATION CONTRADICT THE CONSTANCY OF THE SPEED OF LIGHT - MMXIII
By
Thorntone E. “Butch” Murray
Houston, Texas, USA
June 4, 2013
CONTENTION:
This contention is in strict adherence to and based exclusively on principles presented in the 1905 Theory of Special Relativity (SR) by Albert Einstein. The following mathematical analysis of an ordinary, unremarkable circumstance proves that time dilation and length contraction as presented in SR contradict a fundamental assumption of SR that the speed of light is constant and the same for all observers.
There are very good reasons for physicists such as myself are not willing to spend hours trying to sort out confusing explanations such as this and that's that we've not only done this many times in all our careers many other people try the same thing and in each time it always comes down to one of two things
(1) The author is invariably using alternate definitions of things everyone else is with whom Einstein and his successors used, and that's those based on common sense and
(2) The author is invariably doesn't understand Einstein's derivation to begin with.
As such please go back and show us exactly where in Einstein's article he made an error. Please also prove to us that you understand that Einstein knew that in order for the speed of light to be invariant that both space must be contracted and time be dilated.
If you need help please feel free to follow my own proof at
http://home.comcast.net/~peter.m.brown/sr/sr.htm
Specifically;
Time Dilation - http://home.comcast.net/~peter.m.brown/sr/light_clock.htm
Length Contraction - http://home.comcast.net/~peter.m.brown/sr/lorentz_trans.htm
Also since it’s a given that if time and space are not altered then neither is the Galilean transformation and therefore the speed of light is not invariant.
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I apologize for the delay.
Butch
Alancalverd,
Since this relates to the mathematical foundation of SR, the first paragraph of my last post is very probably as clear as it can get without maths. Sorry.
Pmb,
On your web page concerning light clocks you state, in effect, L is the same in both frames S and S’. Although true, technically L=Ct/2 in frame S and L’=Ct’/2 in frame S’ and L=L’. As such, in your Eq2 and Eq3 equations, Ct/2 can substitute for Ct’/2. Then:
Your Eq2:
(Ct’/2)²=L²
Can be restated as:
(Ct/2)²=L²
Your Eq3:
(Ct/2)²=(vt/2)²+(Ct’/2)²
Can be restated as:
(Ct/2)²=(vt/2)²+(Ct/2)²
Equation Eq3 is invalid for values of speed, v, greater than zero. Therefore, equations, factors, formulas etc resulting from or based on this invalid equation are also invalid. That is inclusive of but not limited to the time dilation factor, the length contraction factor and gamma.
Thank you,
Butch
Thorntone E. Murray
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Alancalverd,
Since this relates to the mathematical foundation of SR, the first paragraph of my last post is very probably as clear as it can get without maths. Sorry.
No need to apologise, but my question remains: what experimental result does your theory predict? What actually happens when you do the experiment? I'm competely unapologetic about being an experimental physicist.
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I'm competely unapologetic about being an experimental physicist.
And you shouldn't! I love you folks! :)
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pmb
You write “Although true, technically..”
It should have been written, “True, however, technically…”
I’ve switched to c for the speed of light and not “C” since the lowercase c is standard notation in relativity. You’ve mislabeled the time parameter in both frames. In S the time parameter is tau (the proper time read on the clock) and in S’ the time parameter is t.
For these purposes inertial frame K’ is in motion relative to frame K at a speed >0<C. In K’ time is t’, length perpendicular to the direction of motion is proper length L’ and distance is d’. In K time is proper time t, length perpendicular to the direction of motion is proper length L and distance is proper distance d. You must agree that as long as what is meant is understood, squabbling over terminology is nothing more than a distraction from real issues and a waste of time.
Your confusion with the symbols led you to your erroneous conclusion.
Please show the instance(s) where confusion with the symbols led to an erroneous conclusion or state clearly this is your opinion.
These errors led you a succession of errors in the rest of your “derivation” which was wrong in the end.
Again, please present the succession of errors or state this is also your opinion.
You should have known that tens of thousands of physicists doing this derivation over every single day for a hundred years
Please cite at least one credible reference for this.
tens of thousands of physicists doing this derivation over every single day for a hundred years would have picked u[ an error long before now if one actually existed
What is the time limit to “pick up” an error? If not “picked up” before that time limit does an error cease to be correctable?
experimental errors would have been found a long time ago too.
What is the time limit for this?
Please, let us confine discussions to facts that can be referenced and label opinions as such.
alancalverd,
what experimental result does your theory predict? What actually happens when you do the experiment?
This is not a theory. It elucidates a logic/mathematical error. There are no experiments. There are no predictions.
Thank you,
Butch
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You write “Although true, technically..”
It should have been written, “True, however, technically…”
[/quote]
It’s not technically difficult at all. It’s quite easy to demonstrate this in any laboratory moving at very small speeds (e.g. 1 cm/s) which is consistent with Lorentz contraction and that’s simply the derivation to show that wires which are neutral in one frame are not neutral in another frame. This can be found in the Feynman Lectures – V-II[1] and in A.P. French’s Special Relativity.
I’ve switched to c for the speed of light and not “C” since the lowercase c is standard notation in relativity. You’ve mislabeled the time parameter in both frames. In S the time parameter is tau (the proper time read on the clock) and in S’ the time parameter is t.
You must agree that as long as what is meant is understood, squabbling over terminology is nothing more than a distraction from real issues and a waste of time.
No. I do not agree. For some reason you feel the need to have switched from the notation that I chose to use for the derivation for another convention, perhaps one that you prefer because you’re used to it. When made the change you also made errors that went along with it. When I pointed out to you what your error was, you then thought that the right approach was to get me to switch back to your notation again, that one that you used in the derivation where you made an error.
Your confusion with the symbols led you to your erroneous conclusion.
Your erroneous conclusion is that there is no length contraction, which has been demonstrated by experiment to be correct.
Please show the instance(s) where confusion with the symbols led to an erroneous conclusion or state clearly this is your opinion.
In this case your change in notation led you to make an error in the conversation that led to your mistake.
The experiments constant with Lorentz contraction can easily be found all over the internet in journal articles and by looking at the literature that particle accelerator labs provide. I don’t do anybody’s work when they can do if for themselves.
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This is not a theory. It elucidates a logic/mathematical error. There are no experiments. There are no predictions.
Your attempt does predict that the results of relativity are wrong.
First off there’s no need to make more changes, i.e. S to K, just to keep the coordinates started with on the web page so I’m using S and S’.
T is time as measured in the frame S’
t is the time as measured in the frame S
The mixing of primes is unfortunate and should be avoided. But its more confusing to mist primes with frames in this case and one of them has to be changed since you felt the need to change coordinates in the first place.
From the diagram we can see that the Pythagorean theorem gives us
(ct/2)^2 = (vt/2)^2 + L^2
T is defined so that L = cT/2. Therefore
(ct/2)^2 = (vt/2)^2 + (cT/2)^2
Solve for t to obtain
t = T/sqrt[1 – (v/c)^2]
Therefore time is dilated.
Your argument on Lorentz contraction was based on that. Therefore that derivation is wrong.
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This is not a theory. It elucidates a logic/mathematical error. There are no experiments. There are no predictions.
Your attempt does predict that the results of relativity are wrong.
First off there’s no need to make more changes, i.e. S to K, just to keep the coordinates started with on the web page so I’m using S and S’.
T is time as measured in the frame S’
t is the time as measured in the frame S
The mixing of primes is unfortunate and should be avoided. But its more confusing to mist primes with frames in this case and one of them has to be changed since you felt the need to change coordinates in the first place.
From the diagram we can see that the Pythagorean theorem gives us
(ct/2)^2 = (vt/2)^2 + L^2
T is defined so that L = cT/2. Therefore
(ct/2)^2 = (vt/2)^2 + (cT/2)^2
Solve for t to obtain
t = T/sqrt[1 – (v/c)^2]
Therefore time is dilated.
Your argument on Lorentz contraction was based on that. Therefore that derivation is wrong.
There are plenty of discussions about the experimental results which prove that time dilation and spatial contraction is true. One only need look.
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This is not a theory. It elucidates a logic/mathematical error. There are no experiments. There are no predictions.
And there we have a problem. Experimentally we find that the predictions of einsteinian relativity are correct, so any mathematical derivation that shows otherwise, is wrong.
No need to go into the maths at all. As in all other aspects of science, from fundamental particle physics, through chemistry and mesoscopic engineering, and right up to astronomy, if it doesn't predict what actually happens, it is wrong, and it is up to the person who derived the "proof" to find out why because the rest of us have more profitable work to do.
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It elucidates a logic/mathematical error.
As you learned, there never was such an error as you originally thought that there was.
To assume that there was an error which has been around for a century which has gone unnoticed is unwise to say the least. To think that it was the other thousands and thousands of people who made a mistake and not you is an error in itself. I not of not one instance that such a thing has ever happened mathematically, at least not at this level of math and application. I'm applying Occamz razor here, of course.
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experimental errors would have been found a long time ago too.
What is the time limit for this?
Did you really believe that you could slip that by as a clever response and that we dim-wits wouldn't know how to answer this kind of questions?
Answer - A scientist wouldn't need to ask and a non-scientist wouldn't understand the answer. The availability of the precision required was available at the time the experiment done and the paper published. So the limit is before the time available that the experimental evidence was available and considered for this experiment.
Experimental verification for length contraction is provided here
http://en.wikipedia.org/wiki/Length_contraction#Experimental_verifications
Please, let us confine discussions to facts that can be referenced and label opinions as such.
We have been. What else did you think we were discussing?
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I understand what he is saying. It is that for a length of rod L, if it is contracted to the external observer then the distance that light is seen to travel in that frame, from an external frame, is also contracted. Therefore light does not appear to travel as far in the same amount of time relative to the two frames. What he fails to appreciate is the differing temporal spaces occupied.
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I would advise thinking about this at the Planck scale before going macroscopic. :-)
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Thank you all for your input.
This is an alternate description:
1. Inertial frame K’ is in motion relative to frame K at a speed >0<C.
2. A measuring rod of length y’ is perpendicular to the direction of motion in frame K’. An identical measuring rod of length y is perpendicular to the direction of motion in frame K.
3. Per the Lorentz Transformations y=y’.
4. The speed formula is t=d/v: Time equals distance divided by speed.
5. In frame K’ light of speed C propagates distance y’, the length of the measuring rod in K’, in the time t’, t’=y’/C.
6. In frame K light of speed C propagates distance y, the length of the measuring rod in K, in the time t, t=y/C.
7. t’=y’/C in frame K’
8. Per the Lorentz Transformations:
9. y=y’
10. Replace y’ with y in frame K’
11. t’=y/C in frame K’
12. t=y/C in frame K
13. Both t’ and t equal y/C. Then:
14. t’=t
15. Assessed by an observer in frame K and assessed by an observer in frame K’:
16. y=y’ The two measuring rods are of equal length.
17. C The speed of light is equal in both frames.
18. t’=t The time is equal for light to propagate the equal length of the measuring rods.
Butchmurray
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(Continued from my last reply)
19. In inertial frame K’ a light path that is perpendicular to the direction of motion is equal in length to the measuring rod of length y’ that is perpendicular to the direction of motion in frame K’.
20. Then, the length of the light path in frame K’ is y’.
21. Light propagates the length of the measuring rod in frame K’ in the time t’, then, light also propagates the length of the light path in frame K’ in the time t’ for Ct’.
22. In K’ for the measuring rod and the light path Ct’=y’.
23. In frame K a light path that is perpendicular to the direction of motion is equal in length to the measuring rod of length y that is perpendicular to the direction of motion in frame K.
24. Then, the length of the light path in frame K is y.
25. Light propagates the length of the measuring rod in frame K in the time t, then, light also propagates the length of the light path in frame K in the time t for Ct.
26. In K for the measuring rod and the light path Ct=y.
27. The equation (Ct’)²+(vt)²=(Ct)² is based on a right triangle and the Pythagorean Theorem.
28. Ct is the hypotenuse and vt is the horizontal side of the right triangle.
29. The vertical side of the right triangle is Ct’, the light path of length y’ that is perpendicular to the direction of motion in frame K’. Ct’=y’.
30. Per the Lorentz transformations y=y’.
31. Ct=y and Ct’=y’ then Ct=y’
32. Ct’=y’ and Ct=y’ then Ct’=Ct
33. In the equation (Ct’)²+(vt)²=(Ct)² substitute Ct for Ct’.
34. (Ct)²+(vt)²=(Ct)²
35. The equation is invalid for v>0.
36. As such, the time dilation formula, the length contraction formula, gamma and all else based on the equation are invalid.
Butchmurray
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A splendid conclusion. What experimental result does it explain or predict?
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Is this why my GPS sends me the wrong way up a one way street.
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Jefferyh,
Very probably not.
Alancalverd,
Here is an example that could be considered a thought experiment.
1. Inertial frame K’ is in motion relative to inertial frame K. All lengths are perpendicular to the direction of motion, in the direction y.
2. In frame K a measuring rod and a light path of equal length are perpendicular to the direction of motion.
3. That equal length is 300,000km.
4. A measuring rod and a light path that are identical to those in frame K are perpendicular to the direction of motion in frame K’.
6. Per the Lorentz transformations as presented in SR y’=y.
7. Then, the measuring rod and light path in K’ are equal in length to their counterparts in frame K and equal to each other.
8. Light propagates at the speed of 300,000km per second for all observers.
9. Measured in frame K, light takes one second to propagate the length of the light path in frame K.
10. Measured in frame K’, light takes one second to propagate the length of the light path in frame K’.
11. The speed of frame K’ is .866C relative to frame K.
12. The time dilation factor for .866C is 2.
13. As such, an occurrence in K’ that takes one second of frame K’ time takes 2 seconds of frame K time.
14. Light takes light one second of frame K’ time to propagate the length of the light path in frame K’.
15. One second of frame K’ time equals 2 seconds of frame K time.
16. Then, it takes 2 seconds of frame K time for light in K’ to propagate the length of the light path in frame K’.
17. In 2 seconds, light propagates 600,000km.
18. Relative to length in K the calculated length of the light path in K’ is 600,000km.
19. The light path and the measuring rod in K’ are of equal length. Then relative to length in K, the length of the measuring rod in K’ is 600,000km.
20. The length of the identical measuring rod in frame K is 300,000km.
21. That inequality is in direct conflict with y=y’
22. When time, t’, is dilated in K’ relative to time in K length, d’, of light paths (and so measuring rods of length equal to the light paths) is increased by the same factor because C is constant.
23. Mathematically: C=d’/t’. It is obvious that when t’ is factored d’ must be factored identically to maintain the constancy of C.
24. For y(d) to equal y’(d’), there can be no time dilation. Time (t and t’) must be equal and the same in both frames.
25. Statements 14, 15 and 16 clearly demonstrate that as a direct result of time dilation light takes unequal times to propagate the length of a single light path in K’. That is not possible if the speed of light is constant.
butchmurray
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Your error is in steps 17 onward. The observer in K is aware that K' is moving and therefore corrects for the apparent time dilatation of events in K' space. No problem - GPS satellites do it all the time.
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Please be more specific. Exactly what is wrong with which step.
Thank you,
Butch
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This is as uncomplicated as it gets.
Per SR: y=y’ (length perpendicular to the direction of motion)
Per SR: the speed of light, C, is constant
Inertial frame O’ is in motion relative to inertial frame O at a speed >0<C.
In frame O: C=y/t
In frame O’: C=y’/t’
Per LT: y=y’
Then:
In frame O: C=y/t
In frame O’: C=y/t’
Therefore:
t’=t
Per the laws of physics, time in frame O (t) and time in O’ (t’) are equal when C is constant and y=y’. That is not a theory.
This equation is fundamental to SR:
(Ct’)²+(vt)²=(Ct)²
Substitute t for t’
(Ct)²+(vt)²=(Ct)²
The equation is invalid for v>0.
The time dilation factor, the length contraction factor, gamma and all else derived from the equation are invalid.
Thank you,
Butch
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Per the laws of physics, time in frame O (t) and time in O’ (t’) are equal
only when measured within the relevant frame. Time dilatation as predicted by relativity (i.e. the laws of physics) turns out AFAIK to be exactly as measured by experiment, which is why I'm interested in the experimental predictions of your theory. O(t') = O'(t), but not O(t).
Once you have stated that
Inertial frame O’ is in motion relative to inertial frame O at a speed >0<C
you are out of the realms of Newtonian physics, which only applies in a single inertial frame. O(t') = O'(t), but not O(t) or O'(t') unless v = 0.
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From Dr. A Oct. 28, 2013
Dear Butch,
if one does not specify what a mathematical equation refers to in physical reality, one can prove or disprove anything.
You say:
This Equation:
(Ct’)²+(vt)²=(Ct)²
Solved for t:
t=t’/sqrt(1-v²/C²) Is the time dilation formula.
Well, then it seems that, if it refers to time dilation transformation, is correct.
But about the same equation you also state:
This equation is fundamental to SR:
(Ct’)²+(vt)²=(Ct)²
Substitute t for t’
(Ct)²+(vt)²=(Ct)²
The equation is invalid for v>0.
So, what does your equation refer to? What physics example refers to?
Is it t = t' or is it t=t’/sqrt(1-v²/C²)?
Best,
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Dear Dr. A Oct. 30, 2013
Clarification #1
Within SR the speed of light, C, is constant. Within SR a length that
is perpendicular to the direction of motion is equal in frame O and
frame O’, y=y’. Therefore, t=t’. That means that time in both frames
transpires at an equal rate. Time in one frame is not dilated relative
to time in another frame. Again, t=t’.
Whenever it was stated that t=t’/sqrt(1-v²/C²) is the time dilation
formula, it was qualified with “The equation is invalid for v>0”. All
of the other equations were/are qualified identically.
Thank you,
Butch
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From Dr. A Oct. 31, 2013
Dear Butch,
it seems that for you the equation
" t=t’/sqrt(1-v²/C²) is the time dilation
formula, it was qualified with “The equation is invalid for v>0”."
does not hold.
However, SR uses it to describe the phenomenon of clock retardation, proved experimentally by the mu meson decay when in motion.
Therefore, it seems to be correct and hold for such a phenomenon.
Is it not, as I told you, that equations assume physical meaning when referred to a specific physical phenomenon?
In conclusion, your statement
"Within SR the speed of light, C, is constant. Within SR a length that
is perpendicular to the direction of motion is equal in frame O and
frame O’, y=y’. Therefore, t=t’."
is correct when it refers to the time t taken by a ray of light perpendicular to direction of motion in O, as measured in S, to travel a distance d,
AND
to the time t' taken by a second ray of light perpendicular to direction of motion in O', as measured in S', to travel a distance d' = d.
So, t = t' for the case above and t=t’/sqrt(1-v²/C²) for the case of the mu meson.
No contradiction.
Best,
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Dear Dr.
Inertial frame S’ is in motion relative to frame S at a speed v>0<C.
From the equation (Ct)²=(vt)²+(Ct’)², H. Lorentz and A. Einstein produced the time dilation and length contraction factors, gamma and much more.
The equation is represented by the uppermost right triangle in Fig. 1. In that triangle the length of the horizontal side labeled vt represents the distance frame S’ advanced relative to frame S in the time t. The length of the vertical side labeled Ct’ represents the length of a light path that is perpendicular to the direction of motion in frame S’. The length of the hypotenuse labeled Ct is the distance the light in the light path in frame S’ traversed as seen from frame S.
To the right of that triangle in Fig. 1, the vertical line labeled Ct represents the length of a light path that is perpendicular to the direction of motion in frame S. It is identical to the light path labeled Ct’ in S’.
Light at speed C traversed the light path of length y that is perpendicular to the direction of motion in frame S in the time t. Then, in frame S:
Ct=y
Light at speed C traversed an identical light path of length y’ that is perpendicular to the direction of motion in frame S’ in the time t’. Then, in frame S’:
Ct’=y’
Per SR y’=y. Relative to length y in frame S, length y’ in frame S’ is the same. Substitute y’ with y for the length of the light path in S’. Then, in frame S’:
Ct’=y
In frame S, Ct=y. In frame S’, Ct’=y. Then, Ct=Ct’. The light paths are equal in length.
The equation used by H. Lorentz and A. Einstein:
(Ct)²=(vt)²+(Ct’)²
Substitute Ct’, the light path in S’ with its equivalent, Ct the light path in frame S.
(Ct)²=(vt)²+(Ct)²
The equation is invalid for v>0.
In the lower of the two right triangles in Fig. 1 the vertical side of the triangle, the light path labeled Ct’ which is in frame S’, is replaced with the light path of equal length labeled Ct which is in frame S. The hypotenuse and the vertical side of the right triangle are the same length, Ct, which is invalid.
The right triangle and the equation are invalid. Therefore, the time dilation factor, the length contraction factor, gamma and all else derived from the equation are invalid.
The results of the mu meson experiments have an alternate explanation.
Time Dilation and Length Contraction Derivation per SR:
The equation that is fundamental to SR:
(Ct)²=(vt)²+(Ct’)²
Reorder
(vt)²+(Ct’)²=(Ct)²
Subtract (vt)² from both sides
(Ct’)²=(Ct)²-(vt)²
Simplify
C²t’²=C²t²-v²t²
Simplify
C²t’²=t²(C²-v²)
Divide by C²
t’²=t²(1-v²/C²)
Square root
t’=t*sqrt(1-v²/C²)
t’=t*sqrt(1-v²/C²) Relative to time t in frame S, time t’ in frame S’ is dilated by sqrt(1-v²/C²).
x’=x*sqrt(1-v²/C²) Relative to length x parallel to the direction of motion in frame S, length x’ parallel to the direction of motion in frame S’ is contracted by sqrt(1-v²/C²).
Thank you,
Butch
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The Least Complicated Proof
Inertial frame K’ is in motion relative to frame K at a speed >0<C.
For any light path, at speed C light traverses the light path of length y in the time t, y=Ct.
A light path of length y is perpendicular to the direction of motion in frame K. An identical light path of length y’ is perpendicular to the direction of motion in frame K’.
The seminal equation H. Lorentz and A. Einstein used to produce numerous formulations such as time dilation, length contraction and gamma is:
(Ct)²=(Ct’)²+(vt)²
For the light path in frame K:
y=Ct
For the identical light path in frame K’:
y’=Ct’
Per SR, y’=y. Then:
Ct=y=y’=Ct’ any two of these are equal.
For this purpose:
Ct=Ct’
The seminal equation:
(Ct)²=(Ct’)²+(vt)²
Substitute Ct for Ct’:
(Ct)²=(Ct)²+(vt)²
This seminal equation, which is fundamental to SR, is invalid.
Thorntone Murray - butchmurray
November 17, 2013
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The Least Complicated Proof
That's more like it, and the diagram helps a lot. Now I can finally be bothered to check to see if you're making the same kind of mistake you did in the past.
For the light path in frame K:
y=Ct
For the identical light path in frame K’:
y’=Ct’
Per SR, y’=y. Then:
Ct=y=y’=Ct’ any two of these are equal.
And you're making the same old mistake - you just hid it in complexity for a time, but now that you've made it easy to see what you're on about, it's now possible to find the error in one minute instead of wasting many hours on it.
y' > y
SR does not require y' to be = to y. SR requires a measurement of y performed from frame k to produce a value for y which is the same as a measurement of y' performed from frame k'. SR does not require y and y' to be the same when measured within a single frame.
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Hi David.
Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.
XI. The Lorentz Transformation
Paragraph 4:
A. Einstein clearly states, y’=y and z’=z
Here is the link:
http://www.bartleby.com/173/11.html
Then, it holds:
For the light path in frame K:
y=Ct
For the identical light path in frame K’:
y’=Ct’
Per SR, y’=y. Then:
Ct=y=y’=Ct’ any two of these are equal.
For this purpose:
Ct=Ct’
The seminal equation:
(Ct)²=(Ct’)²+(vt)²
Substitute Ct for Ct’:
(Ct)²=(Ct)²+(vt)²
This seminal equation, which is fundamental to SR, is invalid.
Thanks.
Butch
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Hi David.
Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.
XI. The Lorentz Transformation
Paragraph 4:
A. Einstein clearly states, y’=y and z’=z
Firstly, because you didn't show in your diagram which direction you're moving one of the frames in, I made the mistake of thinking it was vertical (on the basis that the vertical Ct' in one triangle is replaced with Ct in the other). It now appears from reading your text more carefully that you intend the motion of frame k' to be horizontal, so you can disregard my previous answer.
For the light path in frame K:
y=Ct
That would be your isolated vertical line to the right of your diagram.
For the identical light path in frame K’:
y’=Ct’
That would be the sloping line in the triangle to the top left of your diagram which you have incorrectly labelled as Ct even though it is clearly longer than Ct. You then make a second mistake of labelling the vertical line in that triangle as Ct' instead of Ct and run on into an argument that Ct must be equal to Ct' on the basis of these errors.
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This sums up your problem butch. Sorry about the shaky graphic it is all the effort I could muster for this one.
P.S. Being length contracted the dilated frame sees the photon as traveling twice the distance.
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And the reason you made this mistake is that you neglected the momentum of the dilated frame otherwise time would be accelerated in the moving frame rather than slowed down.
Actually that has just given me an insight. At nearer to light speed velocities the mass must generate stronger gravitation and warp the spacetime in the direction of travel. This is why light speed cannot be broken. It is effectively becoming more and more singular as it nears light speed.
This also implies that a black hole can be accelerated to near light speed. This would explain why galaxies appear to be receding at near or exceeding light speed across cosmic distances as they contain super-massive black holes. Interesting.
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David and JeffreyH,
The interest you both have is greatly appreciated.
Nobel laureates Einstein and Feynman were professors at Caltech. I mention that because Caltech in cooperation with the Corporation for Public Broadcasting (CPB) produced the truly amazing physics series “The Mechanical Universe and Beyond”.
Here is the link for the segment that explains the Lorentz transformation as developed by Lorentz and Einstein.
http://www.learner.org/resources/series42.html?pop=yes&pid=611
I guarantee you it will hold your interest and give you a deeper insight on the subject. Please let me know what you think.
Butch
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David and JeffreyH,
The interest you both have is greatly appreciated.
Nobel laureates Einstein and Feynman were professors at Caltech. I mention that because Caltech in cooperation with the Corporation for Public Broadcasting (CPB) produced the truly amazing physics series “The Mechanical Universe and Beyond”.
Here is the link for the segment that explains the Lorentz transformation as developed by Lorentz and Einstein.
http://www.learner.org/resources/series42.html?pop=yes&pid=611
I guarantee you it will hold your interest and give you a deeper insight on the subject. Please let me know what you think.
Butch
Firstly, I already know about Lorentz transformations. Secondly, do you believe I will find some error in relativity by studying this?
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jeffreyH,
You wrote:
And the reason you made this mistake is that you neglected the momentum of the dilated frame otherwise time would be accelerated in the moving frame rather than slowed down.
Here Einstein explains ‘frames’:
http://www.bartleby.com/173/11.html
Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.
XI. The Lorentz Transformation
Paragraph 3:
“A co-ordinate system K then corresponds to the embankment and a co-ordinate system K’ to the train.” – Frames K and K’
Since a co-ordinate system has no mass, THERE IS NO MOMENTUM.
Yet, you state:
Firstly, I already know about Lorentz transformations.
You wrote:
Secondly, do you believe I will find some error in relativity by studying this?
It was clearly stated “Lorentz transformation as developed by Lorentz and Einstein”
So, quite obviously, the answer is NO.
Again, thank you for your interest.
Butch
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jeffreyH,
You wrote:
And the reason you made this mistake is that you neglected the momentum of the dilated frame otherwise time would be accelerated in the moving frame rather than slowed down.
Here Einstein explains ‘frames’:
http://www.bartleby.com/173/11.html
Albert Einstein (1879–1955). Relativity: The Special and General Theory. 1920.
XI. The Lorentz Transformation
Paragraph 3:
“A co-ordinate system K then corresponds to the embankment and a co-ordinate system K’ to the train.” – Frames K and K’
Since a co-ordinate system has no mass, THERE IS NO MOMENTUM.
Yet, you state:
Firstly, I already know about Lorentz transformations.
You wrote:
Secondly, do you believe I will find some error in relativity by studying this?
It was clearly stated “Lorentz transformation as developed by Lorentz and Einstein”
So, quite obviously, the answer is NO.
Again, thank you for your interest.
Butch
It is the train that has the mass and the train is moving. It is the train's frame of reference that is being compared. I never said a frame has mass. The path of light through a moving dilated frame is straightforward. What is the exact problem?
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Please be more specific. Exactly what is wrong with which step.
Thank you,
Butch
quite simply
17. In 2 seconds, light propagates 600,000km.
18. Relative to length in K the calculated length of the light path in K’ is 600,000km.
That calculation is wrong because the observer in each frame knows that the other is moving (because the received clock pulses are out of sync) and therefore applies the necessary relativistic correction to the signals he receives, and calculates 300,000 km . Every subsequent step is therefore incorrect since it begins with an incorrect assumption.
In short, you are assuming relativity is incorrect in order to prove that it is incorrect. All that is necessary to prove that it is correct is to state that each observer knows that the other has an identical clock.
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That calculation is wrong because the observer in each frame knows that the other is moving (because the received clock pulses are out of sync) and therefore applies the necessary relativistic correction to the signals he receives, and calculates 300,000 km . Every subsequent step is therefore incorrect since it begins with an incorrect assumption.
In short, you are assuming relativity is incorrect in order to prove that it is incorrect. All that is necessary to prove that it is correct is to state that each observer knows that the other has an identical clock.
I remember when I first came across relativity many moons ago I too went round in a few circles. It clicks eventually but is so counter-intuitive that it takes a while.
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Here's one I thought about reading this, its the confusion about how the speed of light is constant for all observers, and time dilation.
1. Observer A is stood at point A and shines a packet of light to a far away point C
2. Observer B is on a rocket ship traveling at speed to point B somewhere between point A and C
3. The packet of light will be travelling at c to both observers
4. But if observer B is having time dilation, travels to point B and back to point A, time will have passed slower for him than observer A but if light hasn't got to C by the time he gets back to point A it will arrive at C simultaneously for both observers
5. How is light then constant seeing as B has been monitoring it for less local time than A?
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How is light then constant seeing as B has been monitoring it for less local time than A?
That is a very good question.
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Here's one I thought about reading this, its the confusion about how the speed of light is constant for all observers, and time dilation.
1. Observer A is stood at point A and shines a packet of light to a far away point C
2. Observer B is on a rocket ship traveling at speed to point B somewhere between point A and C
3. The packet of light will be travelling at c to both observers
4. But if observer B is having time dilation, travels to point B and back to point A, time will have passed slower for him than observer A but if light hasn't got to C by the time he gets back to point A it will arrive at C simultaneously for both observers
5. How is light then constant seeing as B has been monitoring it for less local time than A?
If B is still moving, he will determine that C is nearer to A than if he stops at A and then judges the distance.
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B stops at A..
To give an idea of values, that may show a calculation path,
B is travelling at average 1/2c, to a point B 0.5 light years from point A and obsever A, (1 year there and 1 year back) point C is 3 light years away.
So B would take 2 years with time dilation and arrive back at point A noticing a (?) reduced duration of local time (?) and then observe the conclusion of the light signal at point C at exactly the same time as observer A at this set local time exactly 3 years after light signal was emitted. Therefore the speed of that light was faster than c for observer B as his local time is earlier than local time at A.
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How does each of them know that the light has reached C?
How does B know that he is moving? It would appear to him that both A and C are moving.
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B stops at A..
Which means he has to recalculate everything from the frame A perspective and adjust the amount of time that his journey took to two years, the amount that an observer at A throughout measured for it.
The clock for the traveller who went from A to B and back to A would show a journey time of 1.732 years, but this value is only relevant to any calculations done for the frame for the journey from A to B or the frame for the journey from B to A.
How does each of them know that the light has reached C?
They have to guess that bit, unless they reflect it back and wait another three years for confirmation.
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No, it was all with reference to A.
Actually wiht more investigation I found that B was not an inertial frame of reference as it undergoes acceleration.. So I rephrased my question to.
If B is travelling at constant velocity towards A
When A calculated B is going to arrive at that point (without stopping) in 1 year he emits a light signal towards a reciever 1 light year away at point C.
At the exact point B reaches A the reciever detects the light the time of which takes 1 year form A's point of reference, but obviously less time from B's point of reference.
If I considered the ladder paradox, would this mean that the distance between A and C appears smaller to B? So that B can calculate the speed of light to be constant.
So would that mean that the faster you travel, the closer things appear to be to each other? Is this why its hard to find a parking space?
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No, it was all with reference to A.
Actually wiht more investigation I found that B was not an inertial frame of reference as it undergoes acceleration.. So I rephrased my question to.
If B is travelling at constant velocity towards A
When A calculated B is going to arrive at that point (without stopping) in 1 year he emits a light signal towards a reciever 1 light year away at point C.
At the exact point B reaches A the reciever detects the light the time of which takes 1 year form A's point of reference, but obviously less time from B's point of reference.
It would help if you would give points names that are different from the names of observers so that you can spell everything out clearly and avoid calling anything "that point" without clarifying which point "that point" is.
If I considered the ladder paradox, would this mean that the distance between A and C appears smaller to B? So that B can calculate the speed of light to be constant.
So would that mean that the faster you travel, the closer things appear to be to each other?
The faster you travel, the more space appears to contract to match, so yes.
Is this why its hard to find a parking space?
You should be able to calculate the size of the parking space by adjusting for your relative speeds, and they should be easier to fit the car into than they look at first sight. There's probably an app available to help with that. If there isn't though, it might be worth making one as it could become a popular, fun gift for motorists.
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For these purposes inertial frame K’ is in motion relative to frame K at a speed v>0<C. There are two identical light clocks. One is in frame K. One is in frame K’. Both are perpendicular to the direction of motion. The vertical side of the right triangle is perpendicular to the direction of motion. The horizontal side of the right triangle is parallel to the direction of motion.
The Lorentz factor also known as gamma, 1/sqrt(1-(v²/C²)), is common to time dilation and length contraction. It also has innumerable other applications. The equation (Ct)²=(Ct’)²+(vt)², based on a right triangle and the Pythagorean theorem can be used to formulate gamma.
The Equation:
One of the methods to formulate gamma uses the equation:
(Ct)²=(Ct’)²+(vt)²
Solved for t:
t=t’(1/sqrt(1-(v²/C²))) or
t=t’(gamma)
The Right Triangle (Fig. 1):
The horizontal side, vt, is the distance that frame K’, and so the light clock in K’, advanced in the direction of motion relative to frame K at speed v in the time t.
The vertical side, Ct’, is the length of the light clock which is perpendicular to the direction of motion in frame K’. Within K’ a ray of light at speed C propagates the one-way length y’ of the light clock in the time t’, y’=Ct’.
The hypotenuse, Ct, is the path of the light ray in that light clock as seen by an observer in frame K. Seen from frame K the light ray advances vertically in the light clock as the light clock advances horizontally relative to frame K. The length of the hypotenuse is the distance as seen by the observer in K that the light ray at speed C propagates in frame K in the time t.
The Light Clock in Frame K:
A light clock identical to the one in frame K’ is in frame K. It is of length y and is perpendicular to the direction of motion. Within frame K a ray of light at speed C propagates the one-way length y of the light clock in the time t, y=Ct.
The Time Dilation Issue:
The length of the light clock in frame K’ is Ct’, y’=Ct’.
The length of an identical light clock in frame K is y, y=Ct.
Both are perpendicular to the direction of motion.
Per SR y’=y
Then, Ct’=y’=y=Ct
Then Ct’=Ct
Then t=t’
Since t’=t, the time t’ in frame K’ is not dilated relative to the time t in Frame K.
Per SR t=t’(gamma)
Mathematically t=t’
Time dilation as presented in SR is mathematically contradicted.
The Right Triangle Issue (Fig. 2):
Since Ct’=Ct, substitute the light clock in frame K of length Ct for the light clock in frame K’ of length Ct’ as the vertical side of the right triangle. Then the length of the vertical side of the right triangle is Ct. The length of the hypotenuse is Ct. Per the laws of physics, the hypotenuse is the longest side of a right triangle. The hypotenuse is not the longest side of the right triangle. A law of physics is violated. As such, the right triangle used to formulate gamma is invalid.
The Equation Issue:
Again since Ct’=Ct, substitute Ct for Ct’ in the equation from which gamma is derived.
The original equation:
(Ct)²=(Ct’)²+(vt)²
Substitute Ct for Ct’:
(Ct)²=(Ct)²+(vt)²
The equation is unbalanced. A law of physics is violated. As such, the equation used to formulate gamma is invalid.
By virtue of the three issues presented, gamma is invalid. Therefore, time dilation and length contraction are also invalid.
Thorntone Murray
December 8, 2013
Butchmurray
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Per SR y’=y
What is the import of "per SR" here? You have defined y = y' by saying that the clocks are identical.
[Then, Ct’=y’=y=Ct
Then Ct’=Ct
once again, you began by stating that the clocks are identical. So far, this is a tautology.
but you have forgotten the critical phrase in your opening statement
The length of the light clock in frame K’ is Ct’, y’=Ct’.
If K' and K are moving relative to one another, you can't assume that their observations of each other's clocks are going to be the same as their observations of their own clocks. Neither observer sees a length contraction in his own ship because there is no absolute motion, only relative motion. But each sees a length contraction in the other's ship.
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This is how I see it.
P.S. In fact thinking about it the length contraction is not directly proportional to speed but exponential so the length contraction should not be half. Sorry!
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Per SR y’=y means that per SR length perpendicular to the direction of motion does not contract.
Both light clocks are perpendicular to the direction of motion. They do not contract.
Thank you
Butch
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The Simple Proof
An inertial frame is in motion relative to another inertial frame. One of two identical light paths perpendicular to the direction of motion is in each frame.
1. In accordance with SR, a measuring rod (or light path) perpendicular to the direction of motion in one frame and an identical measuring rod (or light path) perpendicular to the direction of motion in the other frame are the same length compared to each other. Then, identical light paths perpendicular to the direction of motion in both frames are the same length compared to each other.
2. In accordance with SR, time in one frame is dilated compared to time in the other frame. One second in one frame is dilated compared to one second in the other frame.
Of course, within each frame the speed of light is 299,792,548 km/s or 299,792,548 kilometers divided by one second. However, relative to each other the speed of light is 299,792,548 kilometers divided by one ‘dilated’ second in one frame and 299,792,548 kilometers divided by one ‘non-dilated’ second in the other frame.
For speed to be the same when length is the same time must be the same. Compared to each other the times are not the same. Compared to each other, time in one frame is not dilated and time in the other frame is dilated due to time dilation as presented in SR.
Therefore, the speed of light perpendicular to the direction of motion in one frame is not the same relative to the speed of light perpendicular to the direction of motion in the other frame. Time dilation directly contradicts the constancy of the speed of light.
Thorntone Murray
January 4, 2014
Butchmurray.
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Looked at from either frame, the light path in the other frame appears longer because it is not perpendicular. Each account is consistent within itself, judging that one frame is not moving and that the other is moving, while the light paths in the moving one are not perpendicular. The two accounts do contradict each other though, so they cannot both be true, but this is ignored in SR because truth is not considered to be a scientific idea.
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Hi David,
I’ll get back to you as soon as I can.
Thanks,
Butch
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Hi David,
I’ll get back to you as soon as I can.
Thanks,
Butch
In one second light has traveled so far that to make any significant difference to your argument you would need to be travelling at relativistic speeds. You need to consider a square plane of dimensions c^2 to appreciate how difficult it is to support your argument.
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Dear Dr. A, --Sent to Dr A on January 9, 2014--
The Lorentz factor or gamma, 1/sqrt(1-(v²/C²)), is common to time dilation and length contraction. It has innumerable applications. The equation (Ct)²=(Ct’)²+(vt)², based on a right triangle and the Pythagorean theorem, is used to formulate gamma. Fig. 3 details the calculation.
FOR THESE PURPOSES:
Inertial frame K’ is in motion relative to frame K at a speed v>0<C.
There are two identical light clocks. One is in frame K. One is in frame K’. Both are perpendicular to the direction of motion.
The vertical side of the right triangle is perpendicular to the direction of motion.
The horizontal side of the right triangle is parallel to the direction of motion.
DESCRIPTIONS:
The Right Triangle (Fig. 1):
The horizontal side, vt, is the distance that frame K’, and so the light clock in K’, advanced in the direction of motion relative to frame K at speed v in the time t.
The hypotenuse, Ct, is a one-way path of the light ray in the light clock in K’ as seen by an observer in frame K. Seen from frame K the light ray advances vertically in the light clock as the light clock advances horizontally relative to frame K. The length of the hypotenuse is the distance as seen by the observer in K that the light ray at speed C propagates in frame K in the time t.
The vertical side, Ct’, is the length of the light clock which is perpendicular to the direction of motion in frame K’. Within K’ a ray of light at speed C propagates the one-way length y’ of the light clock in the time t’, y’=Ct’.
The Light Clock in Frame K (Fig. 2):
A light clock identical to the one in frame K’ is in frame K. It is of length y and is perpendicular to the direction of motion. Within frame K a ray of light at speed C propagates the one-way length y of the light clock in the time t, y=Ct.
The Equation (Fig. 3):
The formulation of gamma uses the equation:
(Ct)²=(Ct’)²+(vt)²
Solved for t:
t=t’(1/sqrt(1-(v²/C²))) or
t=t’*gamma
ISSUES:
The Time Dilation Issue (Fig. 2):
The length of the light clock in frame K’ is Ct’, y’=Ct’.
The length of an identical light clock in frame K is y, y=Ct.
Both are perpendicular to the direction of motion.
Per SR y’=y
Then, Ct’=y’=y=Ct
Then, Ct’=Ct
Then, t=t’
As t’=t, the time t’ in frame K’ is not dilated relative to the time t in Frame K.
Per SR, t=t’*(gamma). However, mathematically, t=t’.
Time dilation as presented in SR is contradicted mathematically.
The Right Triangle Issue (Fig. 2):
As Ct’=Ct substitute Ct’, the length of the light clock in frame K’, with Ct the length of the identical light clock in frame K as the vertical side of the right triangle used to formulate gamma. Then the length of the vertical side of the right triangle and the length of the hypotenuse are both Ct. The hypotenuse is not the longest side of this right triangle. Per the laws of physics, the hypotenuse is the longest side of a right triangle. As such, the right triangle used to formulate gamma is invalid.
The Equation Issue:
Again, as Ct’=Ct, substitute Ct for Ct’ in the equation from which gamma is formulated.
The original equation:
(Ct)²=(Ct’)²+(vt)²
Substitute Ct for Ct’:
(Ct)²=(Ct)²+(vt)²
The equation used to formulate gamma is unbalanced and therefore, invalid.
From your expert perspective, are these issues substantive?
Thank you VERY much,
Butch
Thorntone Murray
January 9, 2014
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Dear Dr. A, --Reply to Dr. A sent on January 13, 2014--
It will take me longer to respond now than in our previous exchanges. I must ensure that I thoroughly understand your questions and give you the best possible answers.
Your query:
---It seems to me that your issues do not make sense.
Let us examine for example The Time Dilation Issue (Fig. 2). From the figure, according to Pitagora, we have (Ct)²=(Ct’)²+(vt)². So why do you state that "The length of an identical light clock in frame K is y, y=Ct "? Obviously, y = Ct' and not Ct !---
Part 1:
FROM Dr. A: ---The Time Dilation Issue (Fig. 2). From the figure, according to Pitagora, we have (Ct)²=(Ct’)²+(vt)²---
ANSWER: It appears you are referring to Fig. 3, Formulation of gamma. Fig. 1 and Fig. 2 are each a single triangle on one page. The entirety of the other page is Fig. 3.
Part 2:
FROM Dr. A: ---So why do you state that "The length of an identical light clock in frame K is y, y=Ct---
ANSWER: This is specified in DESCRIPTIONS: The Light Clock in Frame K (Fig. 2): (repeated below)
Part 3:
FROM Dr. A: ---Obviously, y = Ct' and not Ct---
ANSWER 1: You are correct for the formulation of gamma and Fig. 3.
Otherwise:
ANSWER 2: It is specified in DESCRIPTIONS: (repeated below) “The vertical side, Ct’ ” and The Light Clock in Frame K (Fig. 2): that:
y’=Ct’ and y=Ct
Per SR:
y’=y
Then, mathematically:
y=Ct’ and y=Ct and y=y’
Repeated for reference:
DESCRIPTIONS:
The vertical side, Ct’, is the length of the light clock which is perpendicular to the direction of motion in frame K’. Within K’ a ray of light at speed C propagates the one-way length y’ of the light clock in the time t’, y’=Ct’.
The Light Clock in Frame K (Fig. 2):
A light clock identical to the one in frame K’ is in frame K. It is of length y and is perpendicular to the direction of motion. Within frame K a ray of light at speed C propagates the one-way length y of the light clock in the time t, y=Ct.
Thank you,
Butch
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Looked at from either frame, the light path in the other frame appears longer because it is not perpendicular. Each account is consistent within itself, judging that one frame is not moving and that the other is moving, while the light paths in the moving one are not perpendicular. The two accounts do contradict each other though, so they cannot both be true, but this is ignored in SR because truth is not considered to be a scientific idea.
It is ignored in SR because there is no such thing as "moving" or "not moving". Objects move (or not) relative to one another: there is no universal frame of reference.
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Hi David. Picture this:
You are on a cruise ship. Your cabin is below decks and has a porthole. The ship will anchor at an island soon. You take a nap. A very loud foghorn wakes you from your slumber. You look out of your porthole and all you can see is the hull of a supertanker moving from your left to your right as you look at it.
You have a long pole with chalk on one end. You stick it out of your porthole and attempt to draw a vertical line on the tanker as it passes by. You start at the top and draw your line straight down. When you are done you can see the line you drew on the tanker is not vertical it is a diagonal line which goes down and to the left.
It is only after the tanker moves out of the way that you tell which boat was moving. If you see you are not moving relative to the island, the tanker was in relative motion (relative to you and the island). If you see the tanker was not moving relative to the island, then your ship was in motion relative to the tanker and the island.
Also, if someone on the tanker drew a line on your ship the same way you drew your line, the lines would be exactly the same.
Butch
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It is ignored in SR because there is no such thing as "moving" or "not moving". Objects move (or not) relative to one another: there is no universal frame of reference.
Which destroys the very mechansim by which things supposedly work. If you want to understand the contradictions, see http://cosmoquest.org/forum/showthread.php?147499-Two-beefs-with-SR-%28special-relativity%29 (http://cosmoquest.org/forum/showthread.php?147499-Two-beefs-with-SR-%28special-relativity%29).
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It is ignored in SR because there is no such thing as "moving" or "not moving". Objects move (or not) relative to one another: there is no universal frame of reference.
Which destroys the very mechansim by which things supposedly work. If you want to understand the contradictions, see http://cosmoquest.org/forum/showthread.php?147499-Two-beefs-with-SR-%28special-relativity%29 (http://cosmoquest.org/forum/showthread.php?147499-Two-beefs-with-SR-%28special-relativity%29).
You do realize because of the extreme speed of light a light path would be as near as damn it vertical at the scales we are used to. No significant angular deflection would be observed. At relativistic speeds the time dilation and length contraction factors actually balance out all the elements of the system anyway. Gravity and momentum are connected implicitly.
Imagine you are moving at 10 miles an hour. The frequency of the gravity waves you would you encounter from the surrounding universe be much less than at relativistic speeds. You are then moving though the gravitational field at huge velocities.
Unless you are taking into account all the factors you can argue black is white and appear to be right.
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You do realize because of the extreme speed of light a light path would be as near as damn it vertical at the scales we are used to. No significant angular deflection would be observed. At relativistic speeds the time dilation and length contraction factors actually balance out all the elements of the system anyway. Gravity and momentum are connected implicitly.
I'm not clear as to how any of that is relevant.
Imagine you are moving at 10 miles an hour. The frequency of the gravity waves you would you encounter from the surrounding universe be much less than at relativistic speeds. You are then moving though the gravitational field at huge velocities.
Likewise, I can't see the relevance in that either, but if you can see a point at which my thought experiment actually breaks down, I'd like to find out what it is. If moving at huge velocities through a gravitational field is possible within SR/GR when there's no preferred frame to tie your gravitational field to, then I'd like to hear how it's done.
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What if the speed of light is directly proportional to the distance you are from the center of the universe. All of a sudden we don't have a universe that is expanding at an ever increasing speed. It certainly is 186,000 mps in our neighborhood, but what about elsewhere?
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A copy of my paper on this matter is located in New Theories with the subject title:The Time Dilation Factor Oversight
Thank you,
Butch