Strange Claims About Relativistic Time Split From A Alternative Theories Thread

[xinhangshen]

Sluggo, you have stated many situations but just restate that Einstein is correct without much reasoning. Everything is just looping back and forth.

 

Now I would have a thought experiment to make the situation more understandable. Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer at any time of that observer's clock. Now observer A sends the image of clock A to observer B. When observer B receives the image, he will adjust his clock to the time of clock A plus the time for the image to travel from A to B. This traveling time will be calculated based on the position of observer B in the reference frame of observer A divided by the speed of light. Then immediately observer B sends the image of clock B back to observe A. Observer A will find that clock B's time is his clock's time minus the time for the image traveling from B to A. This procedure can repeat until both clocks are always synchronized in the frame of observer A.

 

Now let's have a look at the case in the frame of observer B. Once they are synchronized in the frame of observer A, the procedure is completely symmetric relative to both observers. Therefore, the two clocks are perfectly synchronized relative to both reference frames.

Edited August 29, 2016 by xinhangshen

[A-wal]

You are just grumbling for nothing. Please clear your thoughts before making any comments! Stop your nonsense please! If you want to refute my points, please first read my paper and understand my points. Then I will respond to your questions!

I'm not going to ask you questions about a silly BS model if you can't offer anything to support it! Stop trying to claim I'm not using 'reasoning', stop trying to avoid anything that invalidates your stupid model and ANSWER THE QUESTIONS!

 

I'll try again. There's three models.

 

1. A fixed aether.

In this model there can be no test that shows light being constant.

This is completely ruled out by observations.

 

2. A dynamic aether (like sound moving through air).

In this model the only way tests could show a constant speed of light is if the observer always has the same velocity relative to the aether during the tests.

This would require such an unlikely set of conditions to match observations that it can be confidently ruled out.

 

3. No aether.

In this model light always has a constant velocity.

This model matches observations.

 

I assume you're not trying to claim model 1 is the correct one. So if you think model 2 is correct then...

 

1. Why have you tried to push a very old model that was ruled out for very good reasons as if it's something new?

2. Why should a model that's more complicated but explains no more than SR be reconsidered?

3. Why should a model with no evidence and one that needs a very unlikely set of conditions to match observations be reconsidered?

4. You claimed that there's a logical inconsistency in SR. Why have you not been able to point out what it is?

[xinhangshen]

I'm not going to ask you questions about a silly BS model if you can't offer anything to support it! Stop trying to claim I'm not using 'reasoning', stop trying to avoid anything that invalidates your stupid model and ANSWER THE QUESTIONS!

Please behave civilized!

 

 

2. A dynamic aether (like sound moving through air).

In this model the only way tests could show a constant speed of light is if the observer always has the same velocity relative to the aether during the tests.

This would require such an unlikely set of conditions to match observations that it can be confidently ruled out.

How can you call such a simple assertion a logical reasoning? 

 

 

1. Why have you tried to push a very old model that was ruled out for very good reasons as if it's something new?

2. Why should a model that's more complicated but explains no more than SR be reconsidered?

3. Why should a model with no evidence and one that needs a very unlikely set of conditions to match observations be reconsidered?

4. You claimed that there's a logical inconsistency in SR. Why have you not been able to point out what it is?

The aether model is not what I like it or not. It is simply a natural conclusion from all the experiments such as Fizeau experiment. STR is just failed to explain these experiments.

[A-wal]

Now I would have a thought experiment to make the situation more understandable. Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer at any time of that observer's clock. Now observer A sends the image of clock A to observer B. When observer B receives the image, he will adjust his clock to the time of clock A plus the time for the image to travel from A to B. This traveling time will be calculated based on the position of observer B in the reference frame of observer A divided by the speed of light. Then immediately observer B sends the image of clock B back to observe A. Observer A will find that clock B's time is his clock's time minus the time for the image traveling from B to A. This procedure can repeat until both clocks are always synchronized in the frame of observer A.

 

Now let's have a look at the case in the frame of observer B. Once they are synchronized in the frame of observer A, the procedure is completely symmetric relative to both observers. Therefore, the two clocks are perfectly synchronized relative to both reference frames.

What the hell is this. :) I missed that. Oh dear, you really don't understand the first thing about sr do you.

 

 

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer at any time of that observer's clock. Now observer A sends the image of clock A to observer B. When observer B receives the image, he will adjust his clock to the time of clock A plus the time for the image to travel from A to B. This traveling time will be calculated based on the position of observer B in the reference frame of observer A divided by the speed of light.

 

Okay, A sends an image of their clock at midnight and it takes a hour to reach B. B sets their clock to 1:00am, B's clock now shows what A's clock showed an hour ago in B's frame of reference.

 

 

Then immediately observer B sends the image of clock B back to observe A. Observer A will find that clock B's time is his clock's time minus the time for the image traveling from B to A.

 

B then sends the 1:00am message back to A. It takes two hours to reach A from A's frame of reference.

 

 

This procedure can repeat until both clocks are always synchronized in the frame of observer A.

 

No it can't! Once their clocks are synchronised in the frame of A they'll immediately fall out of synch due to the fact that B's clock is running slower than A's clock in A's frame of reference.

 

 

Now let's have a look at the case in the frame of observer B. Once they are synchronized in the frame of observer A, the procedure is completely symmetric relative to both observers.

 

They can never be synchronised for both perspectives because in both frames the other clock is running slower than their own clock. Yes it's perfectly symmetric and will be until one of them accelerates.

 

 

Therefore, the two clocks are perfectly synchronized relative to both reference frames.

 

Therefore you don't even understand basic relativity! You can't simply ignore relativistic effects and then claim that shows that they don't exist!

 

 

Please behave civilized!

Please stop being such a tit then!

 

 

    2. A dynamic aether (like sound moving through air).

 

    In this model the only way tests could show a constant speed of light is if the observer always has the same velocity relative to the aether during the tests.

 

    This would require such an unlikely set of conditions to match observations that it can be confidently ruled out.

 

How can you call such a simple assertion a logical reasoning?

This model treats the propagation of light through a supposed dynamic medium in the same way that sound propagates through air. In order for observations to indicate that sound has a constant velocity all experiments would need to be performed with the observation equipment moving at the same velocity relative to the surrounding air in every test.

 

In order for model 2 to be the correct description of the propagation of light it would require that all experiments would have had to have be performed with the observation equipment moving at the same velocity relative to the surrounding aether in every test.

 

 

The aether model is not what I like it or not. It is simply a natural conclusion from all the experiments such as Fizeau experiment. STR is just failed to explain these experiments.

"The Fizeau experiment was carried out by Hippolyte Fizeau in 1851 to measure the relative speeds of light in moving water. Fizeau used a special interferometer arrangement to measure the effect of movement of a medium upon the speed of light.

 

According to the theories prevailing at the time, light traveling through a moving medium would be dragged along by the medium, so that the measured speed of the light would be a simple sum of its speed through the medium plus the speed of the medium. Fizeau indeed detected a dragging effect, but the magnitude of the effect that he observed was far lower than expected. His results seemingly supported the partial aether-drag hypothesis of Fresnel, a situation that was disconcerting to most physicists. Over half a century passed before a satisfactory explanation of Fizeau's unexpected measurement was developed with the advent of Albert Einstein's theory of special relativity. Einstein later pointed out the importance of the experiment for special relativity."

 

You're completely full of crap!

[xinhangshen]

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer at any time of that observer's clock. Now observer A sends the image of clock A to observer B. When observer B receives the image, he will adjust his clock to the time of clock A plus the time for the image to travel from A to B. This traveling time will be calculated based on the position of observer B in the reference frame of observer A divided by the speed of light.

 

Okay, A sends an image of their clock at midnight and it takes a hour to reach B. B sets their clock to 1:00am, B's clock now shows what A's clock showed an hour ago in B's frame of reference.

 

 

Then immediately observer B sends the image of clock B back to observe A. Observer A will find that clock B's time is his clock's time minus the time for the image traveling from B to A.

 

B then sends the 1:00am message back to A. It takes two hours to reach A from A's frame of reference.

Since there is no difference between the path from A to B and the path from B to A in the reference frame of A as I described, the time for the image to travel from A to B is exactly the same as from B to A, you just don't have any common knowledge of STR at all! Please stop grumbling for nonsense to waste other people's time! 

Edited August 30, 2016 by xinhangshen

[A-wal]

Of course the time is the same if they send their signals at the same time but you said A sends a signal to B and then once B receives that signal they then send it back to A, at which point the distance between them has increased and it takes longer for the second signal to traverse the greater distance. If they were to both send there signals at the same time (not from the perspective of one of them but from the perspective of an object that sits half way between in a frame where they're moving away in opposite directions at the same velocity) then A and B would both see that the other's clock is running slower than their own.

 

I don't consider highlighting a clueless view of sr posing as some kind of breakthrough as a waste of other peoples time. You're the one posting meaning nonsensical crap about something that you can't understand. All I'm doing is pointing out how stupid it is.

[xinhangshen]

It seems there is a need to further explain the problem of special relativity in my thought experiment: 

 

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame. Now observer A wants to know the time of clock B. For example, at 3 o'clock of clock A, observer A sends the image of clock A to observer B. When observer B receives the image, he will immediately send the image of clock B back to observe A. According to special relativity, observer B will see that the time shown on the image of clock A (3 o'clock) will be earlier than the time of clock B (for example 3:15) minus the time for the image to travel from A to B (for example 10 minutes calculated by the position of clock B in the reference frame of observer A divided by constant c, the speed of light) because of time dilation. Now what will observer A see when the image of clock B arrives at observer A? Observer A will see that the time on the image of clock B (3:15) will be later than the time of clock A (3:20 calculated by the time he sent the image to B plus the time for the images traveling from A to B and from B to A) minus the time for the image traveling from B to A (10 minutes). That is, observer A will see that clock B is faster than his clock, contradict time dilation as predicted by special relativity.

 

One may argue that the time for an image traveling from A to B and the time for an image traveling from B to A are different observed in the frame of observer A and in the frame of observer B. OK, just do whatever modification as you want. After your modification, then what time will be shown on the image of clock B received by observer A? You will find, no matter how you choose the time on the image of clock B, the faster clock will always be faster observed in both reference frames. If they are identical, then there is only one answer: they have the same time observed in both reference frames (i.e. the time shown on the image of clock B must be 3:10 here).

Edited August 31, 2016 by xinhangshen

[A-wal]

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame. Now observer A wants to know the time of clock B. For example, at 3 o'clock of clock A, observer A sends the image of clock A to observer B. When observer B receives the image, he will immediately send the image of clock B back to observe A. According to special relativity, observer B will see that the time shown on the image of clock A (3 o'clock) will be earlier than the time of clock B (for example 3:15) minus the time for the image to travel from A to B (for example 10 minutes calculated by the position of clock B in the reference frame of observer A divided by constant c, the speed of light) because of time dilation.

Yes.

 

Now what will observer A see when the image of clock B arrives at observer A? Observer A will see that the time on the image of clock B (3:15) will be later than the time of clock A (3:20 calculated by the time he sent the image to B plus the time for the images traveling from A to B and from B to A) minus the time for the image traveling from B to A (10 minutes). That is, observer A will see that clock B is faster than his clock, contradict time dilation as predicted by special relativity.

No!

 

One may argue that the time for an image traveling from A to B and the time for an image traveling from B to A are different observed in the frame of observer A and in the frame of observer B.

Yes!

 

OK, just do whatever modification as you want. After your modification, then what time will be shown on the image of clock B received by observer A? You will find, no matter how you choose the time on the image of clock B, the faster clock will always be faster observed in both reference frames. If they are identical, then there is only one answer: they have the same time observed in both reference frames (i.e. the time shown on the image of clock B must be 3:10 here).

No, no, NO! B is time dilated from A's frame of reference and A is time dilated from B's frame of reference!

 

A sends the image of their clock to B and B sends the image of their clock to A at the same time in the frame of reference of C who is half way between them with A and B moving away at the same velocity. The clocks show 3:00 when they reach the other one. They then send an image back and both clocks show 3:15 when the reach the other. What you're failing to take into account is that the observers disagree about the time on their watch when the signal reaches the other observer.

 

In A's frame the signal showing 3:00 reaches B at 3:30 but in B's frame they receive the signal at 3:15. In B's frame the signal showing 3:00 reaches A at 3:30 but in A's frame they receive the signal at 3:15. They then both send the second signal showing 3:15 and each confirms that the other is time dilated.

[xinhangshen]

A-wal, please stop making your comments that are always totally wrong!

[A-wal]

That's some damn fine reasoning you've shown there. :)

[sluggo]

zinhang #86

Sluggo, you have stated many situations but just restate that Einstein is correct without much reasoning. Everything is just looping back and forth.

 

Since the Fizeau experiment, there have been many sophisticated high precision experiments that verify the predictions of SR. This is one source as of 2007.

www.edu-observatory.org/physics-faq/index.html 

Now I would have a thought experiment to make the situation more understandable. Suppose there are

two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer at any time of that observer's clock. Now observer A sends the image of clock A to observer B. When observer B receives the image, he will adjust his clock to the time of clock A plus the time for the image to travel from A to B. This traveling time will be calculated based on the position of observer B in the reference frame of observer A divided by the speed of light. Then immediately observer B sends the image of clock B back to observe A. Observer A will find that clock B's time is his clock's time minus the time for the image traveling from B to A. This procedure can repeat until both clocks are always synchronized in the frame of observer A.

Now let's have a look at the case in the frame of observer B. Once they are synchronized in the frame of observer A, the procedure is completely symmetric relative to both observers. Therefore, the two clocks are perfectly synchronized relative to both reference frames.

 

see "sync signal-3gif
The calibration curves are a constant time wherever they intersect an observer path.
It is simpler if possible to sync the clocks as they pass at the origin, if not;
A and B agree to exchange time signals in succession until each has a round trip time for their signal.
Using the SR clock sync convention:
A assigns B .73 to A .79.
B assigns A 1.08 to B 1.16.
This defines the simultaneity axis (red) for each.
Each sees the other clock running slower, a doppler effect since clocks are frequencies.
Each measures the travel time differently. They will stay synchronized until one changes speed.

Edited September 1, 2016 by sluggo

[xinhangshen]

sync signals-3.gif

Since the Fizeau experiment, there have been many sophisticated high precision experiments that verify the predictions of SR. This is one source as of 2007.

www.edu-observatory.org/physics-faq/index.htmlNow I would have a thought experiment to make the situation more understandable. Suppose there are

see "sync signal-3gif

The calibration curves are a constant time wherever they intersect an observer path.

It is simpler if possible to sync the clocks as they pass at the origin, if not;

A and B agree to exchange time signals in succession until each has a round trip time for their signal.

Using the SR clock sync convention:

A assigns B .73 to A .79.

B assigns A 1.08 to B 1.16.

This defines the simultaneity axis (red) for each.

Each sees the other clock running slower, a doppler effect since clocks are frequencies.

Each measures the travel time differently. They will stay synchronized until one changes speed.

Sluggo, you are simply illustrating Lorentz Transformation with graphs, which does not really touch any real physical objects, and therefore nothing will be found. As I mentioned in my paper, relativistic time does dilate after Lorentz Transformation, but relativistic time is just a variable defined by Lorentz Transformation, which can't be materialized. What I want to say is the physical clock time which is materialized and its image can be captured.

 

 

It seems there is a need to further explain the problem of special relativity in my thought experiment: 

 

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame. Now observer A wants to know the time of clock B. For example, at 3 o'clock of clock A, observer A sends the image of clock A to observer B. When observer B receives the image, he will immediately send the image of clock B back to observe A. According to special relativity, observer B will see that the time shown on the image of clock A (3 o'clock) will be earlier than the time of clock B (for example 3:15) minus the time for the image to travel from A to B (for example 10 minutes calculated by the position of clock B in the reference frame of observer A divided by constant c, the speed of light) because of time dilation. Now what will observer A see when the image of clock B arrives at observer A? Observer A will see that the time on the image of clock B (3:15) will be later than the time of clock A (3:20 calculated by the time he sent the image to B plus the time for the images traveling from A to B and from B to A) minus the time for the image traveling from B to A (10 minutes). That is, observer A will see that clock B is faster than his clock, contradict time dilation as predicted by special relativity.

 

One may argue that the time for an image traveling from A to B and the time for an image traveling from B to A are different observed in the frame of observer A and in the frame of observer B. OK, just do whatever modification as you want. After your modification, then what time will be shown on the image of clock B received by observer A? You will find, no matter how you choose the time on the image of clock B, the faster clock will always be faster observed in both reference frames. If they are identical, then there is only one answer: they have the same time observed in both reference frames (i.e. the time shown on the image of clock B must be 3:10 here).

Now I would like to ask you in the situation as shown in the above: what time will be shown on the image of clock B received by observer A if the two clocks run at exactly the same frequency relative to their own reference frame and were synchronized when they started?

Edited September 1, 2016 by xinhangshen

[sluggo]

Sluggo, you are simply illustrating Lorentz Transformation with graphs, which does not really touch any real physical objects, and therefore nothing will be found. As I mentioned in my paper, relativistic time does dilate after Lorentz Transformation, but relativistic time is just a variable defined by Lorentz Transformation, which can't be materialized. What I want to say is the physical clock time which is materialized and its image can be captured.

 

 

Now I would like to ask you in the situation as shown in the above: what time will be shown on the image of clock B received by observer A if the two clocks run at exactly the same frequency relative to their own reference frame and were synchronized when they started?

Typically drawings can show relations much easier than calculations. The CAD system used provides the calculations with sufficient precision. The paths represent 'real' clocks. If A and B are separating at some random speed, they do not have the same frequency. If converging, each sees the other clock running faster. If diverging, each sees the other clock running slower.

Your description is too cluttered to follow.

[A-wal]

Sluggo, you are simply illustrating Lorentz Transformation with graphs, which does not really touch any real physical objects, and therefore nothing will be found. As I mentioned in my paper, relativistic time does dilate after Lorentz Transformation, but relativistic time is just a variable defined by Lorentz Transformation, which can't be materialized. What I want to say is the physical clock time which is materialized and its image can be captured.

Relativistic time IS physical time! That's the point, and it was formulated to be consistent with observations and has since been verified many times.

 

Now I would like to ask you in the situation as shown in the above: what time will be shown on the image of clock B received by observer A if the two clocks run at exactly the same frequency relative to their own reference frame and were synchronized when they started?

The time shown would be the same as the time on their watch when they received A's image, so 3:10. But they received the message at 3:15 on A's watch (after the time it takes for the light to reach them to be accounted for).

 

You will find, no matter how you choose the time on the image of clock B, the faster clock will always be faster observed in both reference frames. If they are identical, then there is only one answer: they have the same time observed in both reference frames (i.e. the time shown on the image of clock B must be 3:10 here).

Utter nonsense based on a total lack of understanding! It works the same both ways because the situation is symmetric. Both clocks are running slow from the other observer's perspective and there's no contradiction.

[xinhangshen]

Typically drawings can show relations much easier than calculations. The CAD system used provides the calculations with sufficient precision. The paths represent 'real' clocks. If A and B are separating at some random speed, they do not have the same frequency. If converging, each sees the other clock running faster. If diverging, each sees the other clock running slower.

Your description is too cluttered to follow.

Sluggo, you are right that CAD can help people understand many complicated things. Your illustration can indeed help people understand the meaning of Lorentz Transformation, but it is not what we want here.

 

What we want here is to solve the contradiction of special relativity arisen from my thought experiment. If you think my thought experiment has any problem, please point out that problem. 

[xinhangshen]

Relativistic time IS physical time! That's the point, and it was formulated to be consistent with observations and has since been verified many times.

It is not the place for you to swear that you are loyal to Einstein. It's the forum for you to provide logical reasoning to prove that relativistic time is the physical time.

 

 

The time shown would be the same as the time on their watch when they received A's image, so 3:10. But they received the message at 3:15 on A's watch (after the time it takes for the light to reach them to be accounted for).

When the image of clock A reaches observer B, the time shown on the image is 3:00, if clock B shows 3:15, then the image of clock B received by observer A will still be 3:15 while clock A shows 3:20 because the time lapse of clock A is the time for the image of clock A traveling from A to B (10 minutes) plus the time for the image of clock B traveling from B to A (another 10 minutes). Deducting the time for the image of clock B traveling from B to A (10 minutes), clock A should be 3:10 which is earlier than clock B. That is, observer A also see that clock B is faster, contradicting the prediction of special relativity.

[A-wal]

It is not the place for you to swear that you are loyal to Einstein. It's the forum for you to provide logical reasoning to prove that relativistic time is the physical time.

I'm not loyal to Einstein. Special relativity is the only model that works with a constant speed of light which it's been shown to be many times.

 

When the image of clock A reaches observer B, the time shown on the image is 3:00, if clock B shows 3:15, then the image of clock B received by observer A will still be 3:15 while clock A shows 3:20 because the time lapse of clock A is the time for the image of clock A traveling from A to B (10 minutes) plus the time for the image of clock B traveling from B to A (another 10 minutes). Deducting the time for the image of clock B traveling from B to A (10 minutes), clock A should be 3:10 which is earlier than clock B. That is, observer A also see that clock B is faster, contradicting the prediction of special relativity.

At 3:00 on observer A's watch observer A sends an image to observer B.

 

At 3:15 on observer B's watch observer B receives observer A's signal. Observer B then sends an image of their own watch to observer A.

 

At 3:45 (let's say the distance has doubled) on observer A's watch they receive the 3:15 signal from observer B and calculate that it was sent at 3:20 own their own watch.

 

 

And it works the same both ways.

 

 

At 3:00 on observer A's watch observer A sends an image to observer B and at 3:00 on observer B's watch observer B sends an image to observer A. These event a simultaneous from the perspective of observer C who is half way between them with both observers moving away at the same velocity.

 

At 3:15 on observer B's watch observer B receives observer A's signal and at 3:15 on observer A's watch observer A receives observer B's signal. Observer B then sends an image of their own watch to observer A and Observer A sends an image of their own watch to observer B.

 

At 3:45 (let's say the distance has doubled) on observer A's watch they receive the 3:15 signal from observer B and calculate that it was sent at 3:20 own their own watch and at 3:45 on observer B's watch they receive the 3:15 signal from observer A and calculate that it was sent at 3:20 own their own watch.

 

 

If two objects are in motion relative to each other then each will be time dilated from the other observer's perspective. All you've done with this 'thought' experiment is taken a continuous view of each other (which can be thought of as a constant stream of signals to each other) and broken it into isolated signals as if that could somehow change what happens.