Unsolvable paradox?

[logicman]

On YouTube is video

Unsolvable Paradox In Einstein's Theory.

SPAM LINK DELETED

Is it really unsolvable?

Edited November 24, 2019 by Strange

[J.C.MacSwell]

Both clocks would in fact show the other as delayed when they are brought together in this symmetric manner.

This is because they would have disagreed on simultaneity earlier.

Not much of a paradox...

[Strange]

On 11/23/2019 at 5:41 PM, logicman said:

On YouTube is video

Unsolvable Paradox In Einstein's Theory.

Is it really unsolvable?

Of course it isn't unsolvable.

Any so-called "paradox" in relativity is simply something that is unexpected or counter-intuitive. The theory is mathematically consistent (so no real paradoxes are possible) and also tested experimentally and confirmed to be an accurate description of the way the world work.

Presumably, the person who made the video is just ignorant of the theory..

[J.C.MacSwell]

25 minutes ago, Strange said:

Of course it isn't unsolvable.

Any so-called "paradox" in relativity is simply something that is unexpected or counter-intuitive. The theory is mathematically consistent (so no real paradoxes are possible) and also tested experimentally and confirmed to be an accurate description of the way the world work.

Presumably, the person who made the video is just ignorant of the theory..

+1. Generally these pseudo paradoxes stem from mixing special relativity with some good ol' Newtonian "common sense".

They aren't compatible except for non relativistic speed approximations.

[Strange]

It is bizarre that there are still people trying to claim that such a well-established theory is wrong. It makes about as much sense as insisting that there is no such thing as oxygen, and it is just dephlogisticated air. Or claiming that diseases are caused by imbalances in the four humors. Quite, quite mad.

[studiot]

But why do I have to look at a Ytube video to find out what is going on?

[Janus]

Despite the author's claim to the contrary, the acceleration of the two stations is an important factor t the solution.  They try to dismiss it by claiming that everything has had some acceleration in the past.  This would have been a valid argument if they they had limited their thought experiment to a period after all acceleration had been performed. But they didn't.   They accelerated both stations after the stations had synced their clocks.  This means that during the period during which station accelerated, they were making their determination of what was happening to the other's station's clock based on measurements made from a non-inertial accelerated frame.   What this means is that for at least some of that period, they would actually conclude that the other station's clock ran fast. not slow and will have gained time during this period.  Once a station stops accelerating, it will return to being an inertial frame and will measure the other clock as running slow.  By the time they meet, the combination of running fast for a period and running slow for a period will result in the clocks reading the same.

 

Of course, if you tried to point this out to the author of this video, they would just call you a "liar".

[J.C.MacSwell]

7 hours ago, J.C.MacSwell said:

Both clocks would in fact show the other as delayed when they are brought together in this symmetric manner.

This is because they would have disagreed on simultaneity earlier.

Not much of a paradox...

I hadn't seen the first 5 minutes where they were stationary when synced up then accelerated. While they would both calculate the other's clock was delayed relative to when I started watching (after the accelerations of each were complete), Janus's explanation is therefore more complete and correct.

Edited November 24, 2019 by J.C.MacSwell

[logicman]

2 hours ago, Janus said:

They accelerated both stations after the stations had synced their clocks.  This means that during the period during which station accelerated, they were making their determination of what was happening to the other's station's clock based on measurements made from a non-inertial accelerated frame.   What this means is that for at least some of that period, they would actually conclude that the other station's clock ran fast. not slow and will have gained time during this period.  Once a station stops accelerating, it will return to being an inertial frame and will measure the other clock as running slow.  By the time they meet, the combination of running fast for a period and running slow for a period will result in the clocks reading the same.

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further?

8 hours ago, J.C.MacSwell said:

Both clocks would in fact show the other as delayed when they are brought together in this symmetric manner.

This is because they would have disagreed on simultaneity earlier.

But they didn't disagree on simultaneity earlier. That's the point.

7 hours ago, Strange said:

Of course it isn't unsolvable.

Any so-called "paradox" in relativity is simply something that is unexpected or counter-intuitive. The theory is mathematically consistent (so no real paradoxes are possible) and also tested experimentally and confirmed to be an accurate description of the way the world work.

Presumably, the person who made the video is just ignorant of the theory..

Instead of talking relativistic rubbish, try to solve this paradox.

7 hours ago, J.C.MacSwell said:

+1. Generally these pseudo paradoxes stem from mixing special relativity with some good ol' Newtonian "common sense".

They aren't compatible except for non relativistic speed approximations.

This is not about the general. It's about the Kaziuk paradox. 

[J.C.MacSwell]

21 minutes ago, logicman said:

 

But they didn't disagree on simultaneity earlier. That's the point.

They would have disagreed during the accelerations, and at all points up until they came together.

If they had been farther apart at the start, on accelerating they would have disagreed moreso.

So much so that if they were far enough apart, due to their own accelerations they would calculate the others clock to have gone forward in time.

Edited November 24, 2019 by J.C.MacSwell

[logicman]

5 hours ago, studiot said:

But why do I have to look at a Ytube video to find out what is going on?

Because this video is the best form of presentation of this paradox.

2 minutes ago, J.C.MacSwell said:

They would have disagreed during the accelerations, and at all points up until they came together.

If they had been farther apart at the start, on accelerating they would have disagreed moreso.

So much so that if they were far enough apart, due to their own accelerations they would calculate the others clock to have gone back in time.

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further?

[J.C.MacSwell]

Edited: back in time should read forward in time

Accelerating in the opposite direction, when far enough away, could move the others clock back in time

Edited November 24, 2019 by J.C.MacSwell

[logicman]

5 hours ago, Strange said:

It is bizarre that there are still people trying to claim that such a well-established theory is wrong. It makes about as much sense as insisting that there is no such thing as oxygen, and it is just dephlogisticated air. Or claiming that diseases are caused by imbalances in the four humors. Quite, quite mad.

For 1000 years Ptolemy's geocentric model was such a well-established theory. So instead of talking relativistic rubbish, try to solve the Kaziuk paradox.

[Janus]

15 minutes ago, logicman said:

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further?

 

The rate at which either station during acceleration would determine that the other station was running fast depends on the magnitude of the acceleration and the distance between the stations.   A brief acceleration and a long time of traveling at fixed speed before meeting,  means that a station, while under acceleration, had to have a high magnitude of acceleration while very far away from the other station. This combines to cause a much faster tick rate for the other station clock.

Since the distance between the clocks during accleration is a factor, increasing the coasting distance by a factor of ten also increases the rate at which you would determine it as ticking by the same factor even if you don't increase the period of acceleration. You still end up with with the observers on each ship agreeing what their clocks read when they meet.

[J.C.MacSwell]

10 minutes ago, logicman said:

Because this video is the best form of presentation of this paradox.

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. And when the stations flew at a constant speed, then from the viewpoint of observer A, the clock at station B was running slower, and from the viewpoint of observer B, the clock at station A was running slower. As a result everything equalized. So when the stations merged, the clocks showed the same time. Is that so? But the stations accelerated very briefly (during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star). So since from the viewpoint of observer A, clock B was running faster very briefly and was running slower for a very long time, and from viewpoint of observer B, clock A was running faster very briefly and was running slower for a very long time, then how after the stations merged, the clocks can show the same time? And what would the clocks show if the stations accelerated the same and were flying at a constant speed 10 times further?

If they were 10 X further away and went through the same acceleration they would read the others clocks adjustment as 10 X as much, so when they came together they would still synch up. 

I see Janus has covered this.

[Janus]

15 minutes ago, logicman said:

Because this video is the best form of presentation of this paradox.

And it is totally a straw-man argument, as it fails to properly apply the entirety of Relativity to the problem, and tries to disprove Relativity by misrepresenting it.

 

29 minutes ago, logicman said:

For 1000 years Ptolemy's geocentric model was such a well-established theory. So instead of talking relativistic rubbish, try to solve the Kaziuk paradox.

And that model was replaced when better observations were made which required a model that made predictions that matched those observations more closely.

 

If Relativity is ever dethroned, it will because new observations or experiments show it to be lacking.   But this requires real actual experiments or observations that conflict wiht what Relativity predicts.   You simply cannot disprove SR by thought experiment.  It has already passed the internal self consistency test.

[logicman]

3 minutes ago, Janus said:

The rate at which either station during acceleration would determine that the other station was running fast depends on the magnitude of the acceleration and the distance between the stations.   A brief acceleration and a long time of traveling at fixed speed before meeting,  means that a station, while under acceleration, had to have a high magnitude of acceleration while very far away from the other station. This combines to cause a much faster tick rate for the other station clock.

Since the distance between the clocks during accleration is a factor, increasing the coasting distance by a factor of ten also increases the rate at which you would determine it as ticking by the same factor even if you don't increase the period of acceleration. You still end up with with the observers on each ship agreeing what their clocks read when they meet.

So in your opinion the situation looks like this:
From the point of view of observer A, the hand of clock B moved forward relative to clock A by a value depending on the value of acceleration of station B, and from the point of view of observer B, the hand of clock A moved forward relative to clock B by a value depending on the value of acceleration of station A. And the clocks will be always aligned regardless of whether the stations cover in constant speed the distance separating me from my mother-in-law, or they cover a distance equal to the diameter of the Milky Way? Is that so? And what if the stations traveled the distance separating us from the Andromeda Galaxy? Or the other end of the universe? Before you answer, let me remind you that special theory of relativity says that the clock delay depends on the speed and DISTANCE that the clock has overcome.

[Janus]

1 hour ago, logicman said:

So in your opinion the situation looks like this:
From the point of view of observer A, the hand of clock B moved forward relative to clock A by a value depending on the value of acceleration of station B, and from the point of view of observer B, the hand of clock A moved forward relative to clock B by a value depending on the value of acceleration of station A. And the clocks will be always aligned regardless of whether the stations cover in constant speed the distance separating me from my mother-in-law, or they cover a distance equal to the diameter of the Milky Way? Is that so? And what if the stations traveled the distance separating us from the Andromeda Galaxy? Or the other end of the universe? Before you answer, let me remind you that special theory of relativity says that the clock delay depends on the speed and DISTANCE that the clock has overcome.

It's not my "opinion" it is what SR concludes when applied in its entirety.

Again, the greater the starting distance, the greater the rate each station concludes the other station clock runs while the station is under acceleration.

If they start light hrs apart, and you spend 1 sec accelerating up to speed, then you will conclude that the other station's clock advanced hrs in that sec.

Start them millions of ly apart, and during that same sec of acceleration you will conclude that the other station clock advanced by millions of years.  That's simply how the proper application of the math of SR works.  No matter how far apart they start, as long as both stations undergo the same proper acceleration for the same period of time, when they meet up , their clocks will read the same according to everyone. 

There is just no way to get SR to disagree with itself.

[Strange]

!

Moderator Note

Moved to Speculations

 

[md65536]

5 hours ago, logicman said:

So in your view, from the viewpoint of observer A, the clock at station B was running faster during acceleration, and from the viewpoint of observer B, the clock at station A was running faster during acceleration. [...] then how after the stations merged, the clocks can show the same time?

There are lots of consistent ways to describe what's going on here. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html (your video is the second half of a twin paradox experiment so all the analyses apply).

 

Your A and B are symmetrical so describing one describes both.

Using Doppler shift analysis, what they observe is: The other's clock appears far behind at the start, and when I move toward it, their clock appears to run faster than mine the entire time until it catches up only when we meet.

If you include relativity of simultaneity in your calculation of the other's rate of time, then depending on your acceleration you can describe it as above: the other runs faster during your acceleration, then slower while at higher velocity.

Or, you can separate time dilation and relativity of simultaneity and say that the other's clock only runs slower the entire time they're moving (as the video suggests), but the acceleration involves a change in inertial reference frame and corresponding change in relative simultaneity, and the clocks are no longer in sync when they start moving. In this case, our clocks are sync'd while we're at rest, but when I accelerate towards you, events where I am are now simultaneous with events at the other clock where the other clock is far ahead of mine, and I spend the rest of the trip catching up to it.

 

Any analysis works and they're all consistent, but to see it (and believe it) you're probably going to have to do the maths. You haven't calculated anything of what A and B would observe, and yet you're finding imagined paradoxes that aren't in the calculations. So start calculating!, and you'll find exactly where the problem lies (where the results don't add up) and then you'll be able to learn the missing pieces that solves the paradox. If it's your first time with the maths, keep it simple and consider only two periods of inertial motion (when they're at rest, and when they're approaching each other after a quick acceleration). I'm sure you'll get lots of help at every step. Just remember there are a lot of different ways to describe it, and you might not believe they're consistent until you see it in your calculations.

 

[studiot]

10 hours ago, logicman said:

15 hours ago, studiot said:

But why do I have to look at a Ytube video to find out what is going on?

Because this video is the best form of presentation of this paradox.

Thank you for replying to my question (Which was actually for the moderators rather than you).

 

What I am not sure of is what youare asking of us.

Originally you asked

18 hours ago, logicman said:

Is it really unsolvable?

 

But then your subsequent posts have been defending the video.

So is the situation that you do not understand Special Relativity and want to learn how it works ?

Or are you saying you understand SR and are promoting the video as correct?

 

Note - the video author clearly doesn't and got his analysis wrong look for instance at this attachment.

 

Can you explain why the velocity in the lower picture is not minus 2x   (-2x) ?

 

It is really important to strip away tha excessive circus and pantomime in the way that video is presented.

So no, I cannot agree that it is the best form of presentation.

Furthermore in your own postings here you have made the mistake of tryign to introduce an absolute frame of reference in saying, for instance

10 hours ago, logicman said:

(during acceleration they traveled a distance equal to the distance separating Earth from the Moon), and they flew at a constant speed for a very long time (they traveled a distance equal to the distance separating the Sun from the nearest star)

 

Distance as measured by whom?

 

The measurement of distance brings in another issue, that of simultaneity.
How do you ensure that the zero on your ruler is at A at the same time you are reading the distance at B?
And what do you mean by this in a world with no absolutes?

This, actually is the prime question the Einstein goes to so much trouble to overcome.

All the suprising effects flow from that.

[Strange]

4 hours ago, studiot said:

So is the situation that you do not understand Special Relativity and want to learn how it works ?

Or are you saying you understand SR and are promoting the video as correct?

Anyone defending the video (which I haven’t watched) would appear not to understand SR

[logicman]

9 hours ago, studiot said:

Thank you for replying to my question (Which was actually for the moderators rather than you).

 

What I am not sure of is what youare asking of us.

Originally you asked

 

But then your subsequent posts have been defending the video.

So is the situation that you do not understand Special Relativity and want to learn how it works ?

Or are you saying you understand SR and are promoting the video as correct?

 

Note - the video author clearly doesn't and got his analysis wrong look for instance at this attachment.

 

Can you explain why the velocity in the lower picture is not minus 2x   (-2x) ?

 

It is really important to strip away tha excessive circus and pantomime in the way that video is presented.

So no, I cannot agree that it is the best form of presentation.

Furthermore in your own postings here you have made the mistake of tryign to introduce an absolute frame of reference in saying, for instance

 

Distance as measured by whom?

 

The measurement of distance brings in another issue, that of simultaneity.
How do you ensure that the zero on your ruler is at A at the same time you are reading the distance at B?
And what do you mean by this in a world with no absolutes?

This, actually is the prime question the Einstein goes to so much trouble to overcome.

All the suprising effects flow from that.

Why isn't -2X in this drawing? Because these are two separate drawings, which were separately in the video earlier, and for preview purposes they were combined into one. These two drawings combined into one are an illustration of the fact that, according to special theory of relativity, from the point of view of observer A station A is at rest and from the point of view of observer B station B is at rest. But apparently this simple obviousness is too complicated for you.

The distances I gave are only meaningless representations of concepts 'far' and 'near'. But apparently this simple obviousness is too complicated for you.

And now I will tell you why I made this entry here. Because I want this video to have as many views as possible. And it doesn't matter what idiots, that can't understand the obvious, are clicking on the video link.

18 hours ago, Janus said:

It's not my "opinion" it is what SR concludes when applied in its entirety.

Again, the greater the starting distance, the greater the rate each station concludes the other station clock runs while the station is under acceleration.

If they start light hrs apart, and you spend 1 sec accelerating up to speed, then you will conclude that the other station's clock advanced hrs in that sec.

Start them millions of ly apart, and during that same sec of acceleration you will conclude that the other station clock advanced by millions of years.  That's simply how the proper application of the math of SR works.  No matter how far apart they start, as long as both stations undergo the same proper acceleration for the same period of time, when they meet up , their clocks will read the same according to everyone. 

There is just no way to get SR to disagree with itself.

Let's say this is the case:

Station A and station B accelerated in the same way to half the speed of light. Then the observers turn off the engines. Now, from the viewpoint of observer A, clock B shows later time than the clock A, and from the viewpoint of observer B, the clock A shows later time.

[It's what you claim.]

When stations are flying at constant speed, from the viewpoint of observer A, the clock B is running slower, and from the point of view of the observer B, the clock A is running slower.

[It's what special theory of relativity says.]

So, according to you, after covering some distance at constant speed by both stations, both clocks will show the same time. Because the relative increase of clocks' running speed would be offset by their relative delay.

Now is the second case:

In the second case everything is the same, exept that the stations covered at constant speed the distance 1000 times bigger. So in this case, relative time dilation was occuring 1000 times longer than in the first case, but the relative increase in clocks' running speed was the same as in the first case. In other words, in the second case the relative time dilation was 1000 times greater than relative gain of time. So even for Beatrice is obvoius that if in the first case the clocks showed the same time, in the second case they couldn't show the same time. But you claim that in every case both clocks would show the same time. When I told it to my she-goat, she almost died of laughter.

14 hours ago, md65536 said:

There are lots of consistent ways to describe what's going on here. See http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html (your video is the second half of a twin paradox experiment so all the analyses apply).

 

Your A and B are symmetrical so describing one describes both.

Using Doppler shift analysis, what they observe is: The other's clock appears far behind at the start, and when I move toward it, their clock appears to run faster than mine the entire time until it catches up only when we meet.

If you include relativity of simultaneity in your calculation of the other's rate of time, then depending on your acceleration you can describe it as above: the other runs faster during your acceleration, then slower while at higher velocity.

Or, you can separate time dilation and relativity of simultaneity and say that the other's clock only runs slower the entire time they're moving (as the video suggests), but the acceleration involves a change in inertial reference frame and corresponding change in relative simultaneity, and the clocks are no longer in sync when they start moving. In this case, our clocks are sync'd while we're at rest, but when I accelerate towards you, events where I am are now simultaneous with events at the other clock where the other clock is far ahead of mine, and I spend the rest of the trip catching up to it.

 

Any analysis works and they're all consistent, but to see it (and believe it) you're probably going to have to do the maths. You haven't calculated anything of what A and B would observe, and yet you're finding imagined paradoxes that aren't in the calculations. So start calculating!, and you'll find exactly where the problem lies (where the results don't add up) and then you'll be able to learn the missing pieces that solves the paradox. If it's your first time with the maths, keep it simple and consider only two periods of inertial motion (when they're at rest, and when they're approaching each other after a quick acceleration). I'm sure you'll get lots of help at every step. Just remember there are a lot of different ways to describe it, and you might not believe they're consistent until you see it in your calculations.

 

Let's say this is the case:

Station A and station B accelerated in the same way to half the speed of light. Then the observers turn off the engines. Now, from the viewpoint of observer A, clock B shows later time than the clock A, and from the viewpoint of observer B, the clock A shows later time.

[It's what you claim.]

When stations are flying at constant speed, from the viewpoint of observer A, the clock B is running slower, and from the point of view of the observer B, the clock A is running slower.

[It's what special theory of relativity says.]

So, according to you, after covering some distance at constant speed by both stations, both clocks will show the same time. Because the relative increase of clocks' running speed would be offset by their relative delay.

Now is the second case:

In the second case everything is the same, exept that the stations covered at constant speed the distance 1000 times bigger. So in this case, relative time dilation was occuring 1000 times longer than in the first case, but the relative increase in clocks' running speed was the same as in the first case. In other words, in the second case the relative time dilation was 1000 times greater than relative gain of time. So even for Beatrice is obvoius that if in the first case the clocks showed the same time, in the second case they couldn't show the same time. But you claim that in every case both clocks would show the same time. When I told it to my she-goat, she almost died of laughter.

[Strange]

2 minutes ago, logicman said:

Why isn't -2X in this drawing? Because these are two separate drawings, which were separately in the video earlier, and for preview purposes they were combined into one. These two drawings combined into one are an illustration of the fact that, according to special theory of relativity, from the point of view of observer A station A is at rest and from the point of view of observer B station B is at rest. But apparently this simple obviousness is too complicated for you.

You seem to have missed the point that velocity is a vector. Therefore, because A is moving in the opposite direction to B, the velocity should have the opposite sign. It doesn't;t matter that they came from different points in your video; the velocities are still in opposite directions.

Also, you need be explicit about which frame of reference the velocity is measured in. (Mixing frames of reference is one of the things that confuses people who don't understand SR.)

 

8 minutes ago, logicman said:

And now I will tell you why I made this entry here. Because I want this video to have as many views as possible. And it doesn't matter what idiots, that can't understand the obvious, are clicking on the video link.

!

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