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Author Topic: Length Contraction and Time Dilation Contradict the Constancy of Light Speed  (Read 18796 times)

Offline jeffreyH

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I would advise thinking about this at the Planck scale before going macroscopic. :-)
 

Offline butchmurray

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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
« Last Edit: 10/10/2013 07:09:30 by butchmurray »
 

Offline 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)+(vt)=(Ct) 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)+(vt)=(Ct) substitute Ct for Ct.

34. (Ct)+(vt)=(Ct)

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
 

Offline alancalverd

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A splendid conclusion. What experimental result does it explain or predict?
 

Offline jeffreyH

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Is this why my GPS sends me the wrong way up a one way street.
 

Offline butchmurray

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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

 

Offline alancalverd

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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.
 

Offline butchmurray

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Please be more specific. Exactly what is wrong with which step.

Thank you,
Butch
 

Offline butchmurray

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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)+(vt)=(Ct)
Substitute t for t
      (Ct)+(vt)=(Ct)

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
 

Offline alancalverd

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Quote
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
Quote
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.
 

Offline butchmurray

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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)+(vt)=(Ct)
Solved for t:
      t=t/sqrt(1-v/C)     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)+(vt)=(Ct)
Substitute t for t
       (Ct)+(vt)=(Ct)
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-v/C)?

Best,

 

Offline butchmurray

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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-v/C) 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
 

Offline butchmurray

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From Dr. A        Oct. 31, 2013

Dear Butch,                   

it seems that for you the equation

" t=t/sqrt(1-v/C) 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-v/C) for the case of the mu meson.
No contradiction.

Best,
 

Offline butchmurray

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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)=(vt)+(Ct), 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)=(vt)+(Ct)
   Substitute Ct, the light path in S with its equivalent, Ct the light path in frame S.
      (Ct)=(vt)+(Ct)
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)=(vt)+(Ct)
   Reorder
      (vt)+(Ct)=(Ct)
   Subtract (vt) from both sides
      (Ct)=(Ct)-(vt)
   Simplify
      Ct=Ct-vt
   Simplify
      Ct=t(C-v)
   Divide by C
      t=t(1-v/C)
   Square root
      t=t*sqrt(1-v/C)

t=t*sqrt(1-v/C) Relative to time t in frame S, time t in frame S is dilated by sqrt(1-v/C).

x=x*sqrt(1-v/C) 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-v/C).

Thank you,
Butch
 

Offline butchmurray

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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)=(Ct)+(vt)

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)=(Ct)+(vt)
Substitute Ct for Ct:
      (Ct)=(Ct)+(vt)

This seminal equation, which is fundamental to SR, is invalid.

Thorntone Murray - butchmurray
November 17, 2013


 
 

Offline David Cooper

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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.

Quote
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.
 

Offline butchmurray

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Hi David.

Albert Einstein (18791955).  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)=(Ct)+(vt)
Substitute Ct for Ct:
      (Ct)=(Ct)+(vt)

This seminal equation, which is fundamental to SR, is invalid.

Thanks.
Butch
 

Offline David Cooper

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Hi David.

Albert Einstein (18791955).  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.

Quote
For the light path in frame K:
      y=Ct

That would be your isolated vertical line to the right of your diagram.

Quote
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.
« Last Edit: 20/11/2013 21:24:57 by David Cooper »
 

Offline jeffreyH

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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.
« Last Edit: 21/11/2013 02:02:57 by jeffreyH »
 

Offline jeffreyH

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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.
« Last Edit: 21/11/2013 02:26:04 by jeffreyH »
 

Offline butchmurray

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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
 

Offline jeffreyH

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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?
 

Offline butchmurray

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jeffreyH,
You wrote:
Quote
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 (18791955).  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:
Quote
Firstly, I already know about Lorentz transformations.

You wrote:
Quote
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
 

Offline jeffreyH

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jeffreyH,
You wrote:
Quote
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 (18791955).  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:
Quote
Firstly, I already know about Lorentz transformations.

You wrote:
Quote
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?
« Last Edit: 24/11/2013 14:21:35 by jeffreyH »
 

Offline alancalverd

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Please be more specific. Exactly what is wrong with which step.

Thank you,
Butch


quite simply

Quote
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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