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Author Topic: Length Contraction and Time Dilation Contradict the Constancy of Light Speed  (Read 18893 times)

Offline jeffreyH

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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.
 

Offline SimpleEngineer

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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?
 

Offline butchmurray

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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.
 

Offline David Cooper

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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.
 

Offline SimpleEngineer

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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.
 

Offline alancalverd

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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.
 

Offline David Cooper

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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.
 

Offline SimpleEngineer

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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?

 

Offline David Cooper

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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.

Quote
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.

Quote
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.
 

Offline butchmurray

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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-(v/C)), is common to time dilation and length contraction. It also has innumerable other applications. The equation (Ct)=(Ct)+(vt), 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)=(Ct)+(vt)
Solved for t:
      t=t(1/sqrt(1-(v/C))) 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)=(Ct)+(vt)
Substitute Ct for Ct:
      (Ct)=(Ct)+(vt)
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


 

Offline alancalverd

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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.

Quote
[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
Quote
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.
 

Offline jeffreyH

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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!
« Last Edit: 09/12/2013 08:45:15 by jeffreyH »
 

Offline butchmurray

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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
 

Offline butchmurray

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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.
« Last Edit: 04/01/2014 14:34:59 by butchmurray »
 

Offline David Cooper

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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.
 

Offline butchmurray

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Hi David,

Ill get back to you as soon as I can.

Thanks,
Butch
 

Offline jeffreyH

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Hi David,

Ill 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.
 

Offline butchmurray

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Dear Dr. A,      --Sent to Dr A on January 9, 2014--

The Lorentz factor or gamma, 1/sqrt(1-(v/C)), is common to time dilation and length contraction. It has innumerable applications. The equation (Ct)=(Ct)+(vt), 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)=(Ct)+(vt)
Solved for t:
      t=t(1/sqrt(1-(v/C))) 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)=(Ct)+(vt)
      Substitute Ct for Ct:
      (Ct)=(Ct)+(vt)
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
 

Offline butchmurray

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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)=(Ct)+(vt). 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)=(Ct)+(vt)---
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
 

Offline alancalverd

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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.
 

Offline butchmurray

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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
 

Offline David Cooper

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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.
 

Offline jeffreyH

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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.

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.
« Last Edit: 18/01/2014 05:48:49 by jeffreyH »
 

Offline David Cooper

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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.

Quote
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.
 

Offline gcrisp

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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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